WOW! A Most Amazing Answer
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- čas přidán 31. 03. 2019
- It's not so simple to solve for this simple answer. Thanks to all patrons! Special thanks this month to: Richard Ohnemus, Michael Anvari, Shrihari Puranik, Kyle.
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I have to stop pretending I will understand any of your videos and just go on with my life and watch cat videos
😂😂😂💔
😂
I spent all day trying to solve this for myself. The solution didn't dissapoint me
Yes
I took 2 min 😏
@@ANDROIDPOSTMORTEM i touqe 1 sec. 😎
@@user-mz7cn9hq8v Great 🔥
@@teamfire3188 What an incredible revelation, so glad you caught them in the act! 😳
I took one look at it and noticed it was the exact same form as the cubic formula, solved for p and q in the depressed cubic, and then solved for x. Fun fact: you can easily find the two complex roots using the cube roots of unity, which are a great tool for solving cubics :)
Ohh yeeee the omega thing right?
damn I thought about exactly the same thing
Yeah and we don't even need to solve for other solutions of the cubic because its determinant is positive meaning there's only one real solution!
Teacher: what is 2 - 1?
r/iamverysmart: my time to shine is now
Jan ka magaling.
It's one.
We’ll use the peano axioms (S(x) denotes the successor of x),
2-1
=S(2-S(1))
=S(2-2)
=S(0)
=1
-e^iπ
π²/g
In the MAJORITY of algebra problems answe is 1
or 2
or 0
That is right. It is easy to see if those solutions will work if we first convert the problem into an equation.
Or 42
Or -1
If it's addition then 0 if it's multiplication then 1
Or e
There is literally a line that says use real valued root instead
I noticed that too lol
How do I get the real valued root using that line?
😂😂😂
@@DctorSkillz1 click on it
@@SadisticNiles All it did was pause the video. Can you help me?
The subtle and maybe unwilling jab at bluepenredpen at the end.
blackpenredpen*
Jab? I must have missed it. What and when was it?
@@OneWeirdDude blackpenredpen basically made a video saying that we shouldn't trust WolframAlpha after a result to an equation wasn't a real root, however, he didn't click the use principle real root button.
@@dwardoyangy6586 Well, this shows a couple things about WolframAlpha.
1. An expression you think you know the meaning of, might not mean that, to WolframAlpha.
2. There are different ways in WolframAlpha to interpret integer-roots, and you need to know the "code" to get the one you intend.
While I don't recall the nature of the WolframAlpha issue that bprp pointed out (although I do recall he had a video with a title along those lines), I think his point in that video was a good one.
Knowing these other little tricks, though, could have improved the value of that video.
@ Luminica: Did you mean, "unwitting jab"?
Fred
dwardo yangy If that was your take away from the video, then you failed completely to understand the point of the video. Too many fell into the r/iamverysmart trap trying to correct him about how to use WolframAlpha even though he knows how to use it, all without even trying to understand what the video is saying. Most people are foolish because they speak a lot without listening and think they know better. Hence most people write foolish comments such as that one. A perfect example of how the Dunning-Krueger effect works.
This is why I never take commenters on CZcams seriously.
Very nice.
Plus this time the solution is well presented.
Loved it, so satisfying. Manipulation of algebraic expressions is like poetry.
Yes! I did figure this one out, thanks for asking :)
A good way to demonstrate the value of learning to use a mathematical toolbox to solve more complex problems without just typing the problem in and getting the answer. This process reinforces available math knowledge and answers the students question, "What are we learning this for?"(Ex. Binomial Expansions). I am 78 years old, did all of my math learning and computation "Without Computer" and taught Math in Secondary Schools for 35 years.
I love math so much, but man that made my brain hurt. Yet it's so weird to see something that looks so complex have such a simple, elegant numerical answer.
I really like your videos and all your content
Thank for sharing this with us my friend
If there were "no computers allowed" no one would be able to watch this video.
What about phones
Rafcio Pranks phones r basically small computers.
@@rafciopranks3570 Phones are computers
- "If there were "no computers allowed" no one would be able to watch this video."
In other words: if you resumed the video in order to figure out the answer, then you were cheating.
Doesn't sound inconsistent to me.
@@yurenchu This is obviously a Zen problem: What is the sound of one hand clapping for a youtube video that you watched without any electronic device?
Beautiful !
Even though cubing seemed at first sight to produce a big mess it also seemed to me to be the only way to tackle this problem.
So I grabbed a pen and paper and started cranking the wheel. Things started to unfold perfectly so I knew I was in the right path.
I usually go with this kind of ideas: 1. Guess a perfect cube: (1+√21)^3 = 1+3√21+3*64*21+21√21=64+24√21=8(8+3√21)
2. Hence (1+√21)^3/2^3=8+3√21 and you easily take a cubic root of the perfect cube.
3. Same thing with the other term.
4. profit
Good one.
Good one. After cubing 1+sqrt(21), I managed to miss that 8 can be factored out (facepalm :) )
This approach also helps when there is only one cube root given and the task is to get rid of it. Such a problem is (much) harder, because simply cubing it gets you nowhere.
Seeing Presh's problem I set out to solve a general case of finding perfect cube given a sum of a rational number and an irratonal number. I arrived at tricubic equation, i.e., cubic equation in terms of x^3. Didn't have the mood to go further... :)
Hello! I'm loving the channel, as it reminices me of my old "Math Olimpics" days. I used to try hard to get the answer right and I had some really good years. Particularly, I remember one problem to which I stumbled upon the answer but I never knew how I got there. The problem is simple and it said "Find the 3 digit number abc, where (abc)-(cba) equals (abc) in any order. Basically, the same 3 digits have to be used. Good luck and see you next time!
- "Find the 3 digit number abc, where (abc)-(cba) equals (abc) in any order. "
The trivial solution is 000 : because 000 - 000 = 000 .
In a decimal system, the only (non-trivial) solution is 954 - 459 = 495 .
In the octal system (= base 8), there are two solutions:
572 - 275 = 275
743 - 347 = 374
The thing that jumps out at me is that their product simplifies nicely.
Let a = (8 + 3 * sqrt(21))^1/3, b = (8 - 3 * sqrt(21))^1/3, then we have:
a^3*b^3 = (8 + 3 * sqrt(21)) * (8 - 3 * sqrt(21)) = 64 - 9*21 = -125
which has a real cube root at -5, so ab = -5. We also have
a^3+b^3 = 16
Let's cube the whole formula, as
x^3 = (a + b)^3 => a^3 + 3 a a b + 3 a b b + b^3
Since the sum of a^3+b^3 is 16, we have:
x^3 = 16 + 3 a a b + 3 a b b = 16 + 3 a b (a+b)
Using ab = -5, we have
x^3 = 16 + 3 * -5 * (a + b)
But x = a+b, so we have:
x^3 = 16 + 3 * -5 * x = 16 - 15 x.
With this, we can read off a solution at x = 1. This checks out. There are
other solutions involving the other cube roots of -125, but they are
likely to be messier.
(-1+sqrt(63) i )/2 and (-1-sqrt(63) i )/2
@Tehom,
- " There are other solutions involving the other cube roots of -125, but they are likely to be messier. "
I'll explore them below:
a³b³ = -125
ab = -5 OR ab = 5(1±i√3)/2
a³ + b³ = 8 + 8 = 16
x³ = a³ + b³ + 3a²b + 3ab²
= 16 + 3ab(a+b)
= 16 + 3abx
... with ab = -5 ...
x³ = 16 - 15x
has the solution *x = 1*
(and two other extraneous solutions, x = (-1 ± i√63)/2 ; but those don't satisfy the original expression)
... with ab = 5(1±i√3)/2 ...
x³ = 16 + 3(5(1±i√3)/2)x
x³ = 16 + [ (15/2)(1±i√3) ]x
x³ = 16 + [ (15/2) ± i(15/2)√3 ]x
x³ = 16 + [ (15/2) + i(15/2)√3 ]x
has the solution *x = ( 1 + 3√21 + i(3√7 - √3) )/4*
(and two extraneous solutions, x = (-1+i√3)/2 and x = (1-3√21 - i(3√7+√3))/4 , which don't satisfy the original expression)
x³ = 16 + [ (15/2) - i(15/2)√3 ]x
has the solution *x = ( 1 + 3√21 - i(3√7 - √3) )/4*
(and two extraneous solutions, x = (-1-i√3)/2 and x = (1-3√21 + i(3√7+√3))/4 , which don't satisfy the original expression)
( Note that for each possible value of ab, always one of the three resulting roots satisfies the equation x³ = 1. )
So the three possible outcomes (depending on how the cuberoot function ³√z is defined for negative reals) of the expression
x = ³√(8 + 3√21) + ³√(8 - 3√21)
are:
x = 1
x = ( 1 + 3√21 + i(3√7 - √3) )/4
x = ( 1 + 3√21 - i(3√7 - √3) )/4
[ Instead of investigating the three possible values of ab separately, we could also bypass that, and instead work with a³b³, as follows:
x³ = 16 + 3abx
x³ - 16 = 3(ab)x
... cube both sides ...
(x³ - 16)³ = 27a³b³x³
... insert a³b³ = -125 in righthand-side ...
x⁹ - 3(16)x⁶ + 3(16²)x³ - 16³ = 27(-125)x³
x⁹ - 48x⁶ + 768x³ - 4096 = -3375x³
x⁹ - 48x⁶ + 4143x³ - 4096 = 0
This is a ninth-degree polynomial equation, whose nine roots are the three solutions and six extraneous roots that I mentioned above. This equation can be relatively easily simplified and solved, by substituting u = x³ and recognizing that 4143x³ = 47x³ + 4096x³ . ]
A very nice problem! I enjoyed this - thank you
Man, I hate myself for not thinking like that. Presh, you are awesome. I wish I could read your books.
You: WOW! This is most unexpected answer for a simpe algebra problem!
Me: and yet, I can't solve it. 😂
This was a very nice solution.
Multiply 8+3*sqrt(21) by 8, you get 64+24*sqrt(21). That's a full cube, 1^3+3*sqrt(21)+3*21+21*sqrt(21), which is (1+sqrt(21))^3. So the first term simplifies to 1/2 * (1+sqrt(21)), similarly the second term simplifies to 1/2 * (1-sqrt(21)). Their sum is 1.
Fascinating and well explained. Answer was easy to understand.
I have found it easier to think of it as
(a^(1/3)+b^(1/3))=(a+b)/((a^(1/3)+b^(1/3))^2-3(ab)^(1/3))
x=16/(x^2+15)
which yields the equation x^3+15x-16=0 with x=1 as a real root.
I believe it is the most orthodox to use Cardano's method for this question
what does ^ mean?
@1bikeman OnDaMoV ,
^ means "to the power of"
For example:
3^2 = 9
means
"3 to the power of 2 equals 9"
169^(1/2) = 13
means
"169 to the power of 1/2 equals 13"
@@yurenchu THANKS!!! 👍👍👍😃
@1bikeman OnDaMoV , no problem! :-)
There is a formula for solving cubic equations in the form ax^3 + bx + c = 0, analogous to the Quadratic Equation for degree 2 polynomials. Interestingly, though, the real root (every cubic must have at least one real root) typically comes out of the equation looking like something you see in this video. So how do you simplify the result, if it can be simplified? Simple - you return to the original equation and test for rational solutions. In other words, you go back to square one, then find the simpler form for the real root by trial and error. Great technique, right?
Now, the time spent calculating the answer using the Cubic Equation wasn't a total waste, as you can run a test on the result to see if a simplification is possible. The test is to look at the a +/- b * sqrt(c) terms inside the cube roots. If a^2 - b^2 * c comes out to a perfect cube, the simplification is possible. To use the number in the video as an example: 8^2 - 3^2 * 21 = -125 = (-5)^3, so the number can be simplified.
Thank you for the video! All of you friends super awesome!
I could very easily follow all the steps and understand this simple easy answer.
I could have never come up with these easy steps to get this simple easy answer myself.
The original expression is just the polinomial's coefficents slammed into Cardano's Formula, needing further simplifications.
I got to the x^3 + 15x - 16 = 0 part on my own, but then I froze.
Thank you for explaining the rest of it.
Never heard of the factor theorem?
Play around with 1+sqrt21 & 1-sqrt21. Or multiply and divide the whole expression by 2. Now use the 2 in the numerator in the cube root and get 64+24sqrt21 and 64-24sqrt21
Since you have already played with 1+sqrt21 and 1-sqrt21, you should know that the above larger values are just cubes of the latter. Now add both and divide the the 2 in the denominator
Good one!As always,I love ur videos.
Trying to click 3:30 several times quickly 😭😭😭
I solved it in a different way, that involves solving each cube root individually:
I found that [1/2 * (sqrt(21) + 1)]^3 = 8 + 3sqrt(21), so you can easily take the cube root of it.
The other was similar, [1/2 * ( - sqrt(21) + 1)]^3 = 8 - 3sqrt(21).
Adding those gets: [1/2 * (sqrt(21) + 1)] + [1/2 * ( - sqrt(21) + 1)] = 1/2 * 2 = 1, the answer. Full proof below:
I've seen tricks that sqrt(a+sqrt(b)) can some times be solved if you work backwards, like 13+4*sqrt(7) = 2^2+2*2*sqrt(7)+sqrt(7)^2 = (2+sqrt(7))^2
Therefore, I thought that each cube root would involve some k * [sqrt(a) + b]^3 inside of it, that would be equal to 8+3*sqrt(21).
if you expand that you get k * [(sqrt(a))^3 + 3*(sqrt(a))^2*b + 3*sqrt(a)*b^2 + b^3]
Simplify: k * [a*sqrt(a) + 3*a*b + 3*sqrt(a)*b^2 + b^3]
Disregard the k for a moment, then group rest: sqrt(a) * [a + 3*b^2] + 3*a*b + b^3
We need this to be equal to 8+3*sqrt(21), for the first one. From this, we can conclude a = 21.
Plug it in: k * [ sqrt(21) * [21 + 3*b^2] + 63*b + b^3 ].
k * [21 + 3*b^2] = 3, k * [63*b + b^3] = 8. We can see from the first equation that k must be less than 1, so it will be simpler to solve for a different constant c, equal to 1/k. Then we can move this over to the other side.
21 + 3*b^2 = 3c, 63*b + b^3 = 8c, a system of equations!
It is easy to solve for c on the first equation: 7+b^2 = c
Put in the other equation: 63*b + b^3 = 56 + 8 * b^2, and this is a cubic equation
b^3 - 8b^2 + 63b - 56 = 0. Possible rational roots inculde +/- 1, 2, 4, 7, 8 ... (rational roots theorem)
After testing, 1 is found to be correct. Divide out to see if there are others: (b^2 - 7b + 56)*(b-1)
The determinent of b^2 - 7b + 56 is 49 - 4*56, a negative number, so there are no other real roots.
7+b^2 = c resulting in c = 8, so k = 1/8.
Now taking the original expansion: 1/8 * [ sqrt(21) * [21 + 3*1^2] + 63*1 + 1^3 ] = 1/8 * [24*sqrt(21) + 64]. This is equal to the original expression in the cube root. We now know that this is equal to 1/8 * (sqrt(21) + 1)^3, which we can expand and see that it is correct.
This is easy to take the cube root of: 1/2 * (sqrt(21) + 1). This is the answer to the first cube root.
For the second one, the only thing different is the negative. If we cube 1/2 * ( - sqrt(21) + 1), the only place when it will change is when the -sqrt(21) is raised to an odd power. This is also the only time that you get a sqrt(21) in the end, so adding this negative will reverse the sign on the sqrt(21) at the end, which is what we want.
So now the final answer is:
1/2 * (sqrt(21) + 1) + 1/2 * (-sqrt(21) + 1), sqrts cancel and you are left with 1/2 * 2 = 1, and your done!
I like this better because it solves each one individually, and could be used to solve problems that only have one, in which his method could not be used.
Eyyyy another cuber!
@@AndrewTyberg Hi
This is the second problem on your channel that happened to be similar to my mock AMC 10 I created. Wow, these problems are nice.
Brilliant. Simply brilliant.
It can be shown that (8 + 3*sqrt(21)) = [1/2 + 1/2 *sqrt(21)]^3 and (8 - 3*sqrt(21)) = [1/2 - 1/2 *sqrt(21)]^3. (Though there does not seem to be an easy way to figure this out.) So the answer is simply (1/2 + 1/2) = 1.
BPRP taught this one to me!!
Once you think about substituting the values for a variable, it becomes pretty straightforward
Nice, ideal for solving in mind.
And for this time, I got the answer correctly 😄
How can i ask question from u?
Thanks for the video!
I used to try these problems in my 9th grade. My math teacher - Mr. Anil Kumar - showed me the ideas of how to tackle them.
My reaction to the approach and answer: 3:30
5:02 I could have used a little help at the "then you expand" part.
502 is the end of the video, where he's plugging his books 'n stuff.
"Then we expand" is at 1m13s, and it's just the binomial expansion, with exponent=3; but with the last term moved into the 2nd spot:
(c + d)³ = c³ + 3c²d + 3cd² + d³
= c³ + d³ + 3c²d + 3cd²
. . . with c = ∛a, d = ∛b
Then he factors those last 2 terms:
= c³ + d³ + 3cd(c + d)
Fred
Great! In morning I prefer this over coffee. Thank you.
Excellent!
Hey solve the last equation by synthetic division getting different answers
Maybe you used -1: if the factor is (x-1), use the zero, which is x=1, to divide using synthetic division
I guessed 1 and it is right! *(because you said the answer will be unexpected)*
So you expected it, and then it contradicts the premise.
I am so glad I figured it out in 20 sec. It took me nearly 3 minutes to solve.
You can set x = A+B, where A and B are the two pieces. Then use the identity: (A+B)^3 = A^3+B^3+3AB(A+B). That yields the cubic equation: x^3 = 16 - 15x and then the real solution x=1 by inspection, after confirming that the other roots are complex solutions to: x^2 + x + 16 = 0.
A simpler way to prove that x=1 is the only solution is to take the derivative of the cubic polynomial. It's positive for all x, which means that the curve can only cross the axis at one point: x=1.
The expression looks suspiciously like the solution of a cubic equation. It took me considerable round-about work to arrive at this solution, but: first, multiply the two cube roots. The result is -5. Then call the entire expression x, and cube it: x^3 = 16 + 3x(-5), the second term coming from the cross multiplication. So x^3 + 15x = 16.
The obvious solution is x=1, and it has to be the only real solution, because x^3 + 15x is strictly increasing. Consequently, that complicated expression is 1.
You could have also solved it by applying the (depressed) cubic formula in reverse! And you can tell that there is only one real solution to that equation from the fact that the determinant (the expression under the square roots in the cubic formula) is positive.
@@tahamuhammad1814 Was "depressed" meant in the sense of "deprecated"? This is one example illustrating why the cubic formula might be deprecated: it obscures the basic simplicity of integer solutions, to the point of having to revert to the original cubic to evaluate the solution.
@@JohnRandomness105 A "depressed" cubic is a cubic equation with no quadratic term. And in general, a depressed nth power polynomial is a polynomial with no (n-1)th power term.
@@JohnRandomness105 the "depressed" cubic formula is the formula specific to "depressed" cubics
@@tahamuhammad1814 Just goes to show how familiar I am with cubic culture: I don't know what a "depressed" cubic is. I simply derived my own method for solving them, which became partly formula. Either of these equations appeared:
x³ + 3x = 2A or x³ - 3x = 2A
Either I went the trig/hyperbolic route, or the same thing repeated so many times that I finally memorized the formula for those two equations. (In the second version, when A² < 1, the formula forces me into the trig version.)
I read this in G.N. Berman. It is possible to calculate the cube roots and then add also.
I just sort of estimated the roots and ended up with "somewhere around 1" and decided that made sense since the video title told me it was an interesting solution
I knew that it was going to be 1
I mind my decisions 😁😁
Your videos are good sir.
Such questions may be asked in the category of Integer type questions in Indian entrance exams.
it can be shown that the two cubic roots can be written as (1+sqrt(21))/2+(1-sqrt(21))/2 which yields 1
The original expression suggests the Cardano solution to a cubic. The general cubic a*x^3+b^x^2+c*x+d=0 can be rearranged so as to read x^3+p*x+q=0 - that is the quadratic term is eliminated by a simple substitution. The solution to the latter equation is (-q/2+(q^2/4+p^3/27)^(1/2))^(1/3)+(-q/2-(q^2/4+p^3/27)^(1/2))^(1/3). Substituting the values in the given problem we have : -q/2=8 and (q^2/4+p^3/27)^(1/2)=3*(21)^(1/2) from which it follows the q=-16 and p=15. Thus the required value is the real root of the equation x^3+15*x-16=0 which by inspection we can see is x=1, and the other two roots are imaginary, so the expression as given evaluates to 1.
this looks like it was manufactured to be 1 by simply using the cubic formula, kinda cool I guess...
Rewatched the video very carefully to see if this was an April Fools joke.
That looks exactly like the sort of thing you'd get from the solution formula for the cubic equation. I don't know that formula well enough to try to reproduce the cubic it would have come from.
Also, the second cube-root term is negative: 3√21 = √189 > √64 = 8
Ball-parking it, it looks to me like it's close to 1, and could just be = 1.
Now having watched it, yes that was very well done!
My hunch was correct, but that was only a hunch; it's nice to see exactly how to get that result.
Fred
Really amazing sir this problem please more videos on this topic..
Isn't this just because (8+3sqrt(21))^(1/3)=(1+sqrt(21))/2 and the conjugate equation, so their sum is 1?
I'm convinced you've found the best way to do this, but could you explain it more clearly please?
I can verify that what you wrote is correct by cubing both sides, but I'm curious how you derived it.
FWIW, I solved it using mostly the same technique that Presh did. My derivation was a bit different, but I wound up with the same equation x**3 + 15x - 16 = 0, at which point I basically guessed x=1.
This is a bit Galois-theoretic, so hang on. The first thing to do is to solve it Presh’s way. So you know the answer is 1. Then you look at that and say, hey, why in the world should two weird irrational-looking numbers have an integer sum? And you verify that these two numbers are Galois conjugates. Them along with their third-root-of-unity multiples would be all such conjugates, if there’s nothing fishy going on, and the trace of the elements would be zero since they satisfy (x^3-8)^2-189=0, whose x^5 coefficient is 0. But you usually can’t have two Galois conjugates whose sum is rational unless they’re actually their only conjugates, so you begin to wonder if something fishy is going on.
A way out of the trace debacle is perhaps (x^3-8)^2-189 isn’t really their minimal polynomial, and perhaps it’s reducible. And then you see that you have two of its roots whose sum is 1 and whose product is -5, as Presh calculated. So these two things are actually roots of x^2-x-5, whose roots are the numbers I described. You see this and check that this is actually a factor of (x^3-8)^2-189=x^6-16x^3-125=(x^2-x-5)(x^4+x^3+6x^2-5x+25), and all is right with the world.
And so in fact, if you ever come across a problem like this, where a cube root of something plus a cube root of something else is an integer, there will always have to be some sort of easy answer like this lying beneath it. Galois theory basically says so.
Where this generally appears is in using Cardano’s formula to solve a cubic. The solution form for x^3+ax+b=0 is x=u+v, where uv=-a/3 and u^3+v^3=-b. (Solving these, you get that u and v are cube roots of the roots of some quadratic polynomial.) If you know your solution is nice, then the ugly cube root form Cardano spits out must simplify somehow.
I did it this way, by realising that 8 + 3sqrt(21) = ((1+sqrt(21))^3)/2^3, and 8 - 3sqrt(21) = ((1-sqrt(21))^3)/2^3
Not gonna lie this problem was actually pretty cool.
You get formulas like this as solutions to cubic equations all the time, with them often becoming much simpler.
This means you are studying for IIT. That is how I used to think 25 years ago.
Thanks for this.
the answer is 1....
hint 64-189=-125=-5^3
4:14 or you can use the goddamn "Use the real valued root" button instead ay
Instead of cuberoot you can also write cbrt in WolframAlpha (which is what I did).
Also interestingly, when typing the same formula into HP Prime app, it shows me that ³√(8±3√21) = (1±√21)/2, so when you add both terms the roots will cancel out. (It is easy to prove the latter by squaring both sides, but hard to come up with that form I guess if you are not a computer that can try many options quickly)
Excellent! IOW,
(1 ± √21)³ = 1 ± 3√21 + 3·21 ± 21√21 = 64 ± 24√21 = 8(8 ± 3√21);
1 ± √21 = 2∛(8 ± 3√21)
That's really great - thanks!
Fred
@mihiguy,
If x = ³√(8 + 3√21) + ³√(8 - 3√21), then x is real and there exists a real d such that
x/2 + d/2 = ³√(8 + 3√21)
and
x/2 - d/2 = ³√(8 - 3√21)
After cubing, this results in
(x ± d)³ = (8 ± 3√21)*2³
x³ + 3xd² ± (3x²d + d³) = 64 ± 24√21
from which we can derive
x³ + 3xd² = 64
3x²d + d³ = 24√21
These two independent equations in two unknowns can be combined and solved for x and d, which takes a rather long calculation (and is therefore indeed hard to come up with). However, a keen eye may already notice that
x³ + 3xd² = 64 = 1 + 3*21 , and
3x²d + d³ = 24√21 = 3√21 + 21√21
and hence d = √21 and x = 1 .
Really nice solution!!
For once I was able to solve one of these without watching the video.
I don’t think he realizes the power he has. He is so smart he can make a video where he says 2+2=5, put some random numbers and math, and make us all believe it, just an idea for next April 1st, plz like this so he sees this
You made it visibly hard by introducing a and b.
We have to just cube and take cube root simultaneously.
Then expand it by formula of (a+b)^3.
We will simply get the answer
There is a simpler way to solve this.
Just replace 3*sqrt(21) with sqrt(189) which is similar to sqrt(8^2 + 5^3). Now the expression is similar to the cubic formula.
Just notice that the -(b/3a) term is missing, thus you can deduce that b = 0.
Then you an see that -(d/2a) = 8 and c/3 = 5
Simply chose a=1
=> our solution must solve the following polynomial: x^3 +15x -16 = 0
And it’s easy to see that 1 is the only real solution
Bro TRY JEE ADVANCED MATHS QUESTIONS
@Another Random Cuber isko pata nhi advanced lvl q kaise hai hai apni jee ke exam mein
Either this was a jargon filled April fools prank or I just didn't understand any of it
Nice try, but nobody can fool me 😎😎😎
The best April Fools joke I read about today came from this website
*Read more*
@@JSSTyger nobody's going to fall for it
@@Nylspider you'd be surprised.
JSSTyger, you can't fool me! I'm an old-youtube-layout user ;)
This is not a joke....it is formula based and years and years of redundant training.....it is not much of a riddle.
taking the negative out one or the other of of the radicals will give the other two solutions to the cubic as the 'principle' in wolframalpha . in this case, writing it as:
(8 + 3√21)^(1/3) - (- 8 + 3√21)^(1/3)
will return 1
The answer is 2 times 5 times (2/5) by using hyperbolic functions.
This kind of problems can be easily produced in hyperbolic angle numbers.
I knew it was gonna be 1😂😂😂
Everytime the most complicated methods have the most simple answers
😂😂😂😂
In India, these questions are just high school level
For Cramers only.
Congratulations
that's why students after passing 10+2 go and take JEE coachings.
[don't be modi(fekoo)]
Half population of 10 years old indian kids cant do subtraction.
Substituting variables to preform operations in sequential order easier.
this is so great!!
Can you fix your video titles to be informative? Now you can't tell if this a silly game show answer or what.
@Shravan The difficulty of the puzzles on this channel varies dramatically. And it's hard to know from the thumbnail and title what kind of video you're getting.
@@qwertyTRiG So what? Do you only watch certain MYD videos based on the title?
@@Grizzly01 Not I, but it wouldn't be unreasonable to have a way to select the more interesting ones, which seems to be what OP was requesting.
@Shravan: If you search for math videos, it's kinda hard to find this one with that title.
*sees video in feed*
*types equation into a calculator*
*calculator returns NaN*
*wonders why*
*watches video*
*understands why*
Wonderful genius solution
In High School, I excelled in Math and Science. Almost didn't graduate because I had no desire for English or History. Well to make a long story short, it took me about 10 minutes to just look at the problem and see that it would result in X=1.
Before watching i guess the answer is 1
And this is actually the right answer, so I don't really agree that this is the most unexpected answer :P
@@nijucow cichaj babo
Same
You are exactly correct! It is NOT the most unexpected answer. It is the most AMAZING, as stated in the title.
Did you mean: *The* _most_ amazing answer?
Tom Riddle dude thats -1 lol
Amazing answer👍
To simplify. 3 x sq. Root of 21 = 63. Usually u go lower to find sq. Root, which is 49 = 7sq. 8-7 = 1
The other side, answer 15. He went higher and kept it root. -16 minus 15= 1. The answer lower would be root 9 = 3 sq.
I'm from India and this question was given us in 9th standard 😂😂
And did you get it right?
I also study in india but i think my teachers are cheaters😭😭
@Sekijo: Monkeys Die Twice Don't worry. Europe has many mathematicians like Euler, Guass, Jon von Neumann. India has produced only one brilliant mathematician and he was Srinivas Ramanujan and he wasn't even recognised here , he had to go to Cambridge University in Britain where his talent actually got recognised. The problem is that there are so many laws , formulae and theorem which we learn that we don't need to do any critical thinking to solve a question like this. Just recognise the theorem and just apply . So you'll see a lot of people solving difficult problems but you'll probably never see someone discovering a new theorem or contributing something to pure maths that's not going to happen because the education system simply doesn't encourage that
1:44 Difference of 2 squares 8² -3²(21)- oh and remember from the instructions: No calculators! Okay... so lets see... um... 64... uh... minus.... um... 6 times... 21... that's uh... 180.... and... um 6... so.. 186... um.... minus- No wait... 64... uh... um...
Yeah... """ -5 """ how'd you get there Presh? Did you decide to use a calculator?! Did you fail to mind your decisions?
**mike drop**
EighteenCharacters 3^2 is not 6. 3^2 is 9. 64 - 9*21 = 64 - 189 = -125. The cube root of -125 is -5.
@@angelmendez-rivera351 I think he's just making a joke about how the mental math requires a calculator for the average person (or a piece of paper but who has time for that these days)
Cringe
Before I watched the video without realizing we were talking proofs here. I thought it was surprisingly easy and straightforward. Oh well. By now I should have been aware that Presh doesn't just give a simple math problem without demanding some more.
This is an amazing solution.
This is simple math problem we learned during grade 9 in Asia.
Where in asia?;
Nice April Fools vid
You fooled us that this us going to be one
Dude is not easy term so how can u handel
@@atharansari8579
As long as I can handle your grammar and spelling
@@randomdude9135
I know, genius
@@randomdude9135
You are missing the point, genius. Also, who adds . (period) after a ? (question mark)?
Dude just go to a English studying channel. Maths is not your cup of tea.
Immediately recognised Cardano's Formula for the cubic in the style of the problem and knew there'd be something to do with cubics here.
I'm studying for the Brazilian Navy School admission test for the Middle School first grade, and the kind of question of it is like this. You could take a look in their tests