A Hexic Equation | Can You Solve?

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[@peterchan6082]

x⁶ = (5x - 6)³
(x²)³ - (5x - 6)³ = 0
From the formula for difference of two cubes, we have
(x² - 5x + 6)[x⁴ + x²(5x - 6) +(5x - 6)²] = 0
So the other four roots come from the [quartic in square brackets].

[@scottleung9587]

I got 2 and 3, but the nonreal solutions were a little far-fetched for me.

[@akuchak1]

Hello there,
How about to rewrite the given equation as follows:
(X^2)^3 - (5X -6)^3 = 0
(X^2 - 5X +6)[X4 + X^2(5X-6) + (5X-6)^2] = 0
Either:
X^2 -5X+6=0 => X=2 , X=3
Or:
X^4 + 5X^3 - 6X^2 + 25X^2 - 60X + 36=0
The roots of the recent equation are all complex numbers.
X^4 + 5X^3 +19X^2 - 60X + 36 = 0

[@JefiKnight]

I was expecting him to do something like that.

[@maxhagenauer24]

You can tell right away that it must be greater than 1 because if it were less than 1, it would make what's in the parenthesis negative and cubing that still is negative while on the left hand side, its to an even power to make it positive. So try the next integers up with 2 and 3 and they both work.

[@richardl6751]

If we change this slightly to x^6=|(5x-6)^3| then -6 is also an answer.

[@TedHopp]

A general quintic equation cannot be solved in radicals. However, there is a general solution in terms of square roots, cube roots, and ultraradicals. (The ultraradical of a real number a is the unique real root of the polynomial x^5 + x + a.)
The Wikipedia article "Bring Radical" has a nice writeup, including details on how to solve the general quintic.

[@SyberMath]

Nice but pretty heavy!

[@TedHopp]

@@SyberMath Heavy indeed!

[@Qermaq]

Always fun. Here's an idea that's kinda trigonometric but I think you'd need calculus to solve it because I am getting stuck.
So the diagram is a quarter circle with radius r and center o. A is the vertical radius's intersection with the quarter circle, and B is the horizontal intersection. Construct line CD where C is on AO, D is on the quarter circle and CD is parallel to OB, AND such that the area of region ACD equals the area of region CDOB wihch equals (ipr^2)/8. Call the length of OC a and the length of CD b. Find a and b in terms of r.
If we draw OD we have angle alpha as both DOB and CDO. DOC is a right triangle and we can find its area, ODB is a sector with angle alpha and we can find its area. But then I get stuck at 2*alpha + sin(2*alpha) = pi/2. I can express a and b in terms of alpha easily enough, but in terms of r? That is eluding me....

[@premsah7799]

Can anyone solve this , the max value of Sin^2(Cosx)+Cos^2(Sinx)

[@SyberMath]

www.wolframalpha.com/input?i=+the+max+value+of+Sin%5E2%28Cosx%29%2BCos%5E2%28Sinx%29

[@DonRedmond-jk6hj]

Since you asked ... see Math Stack Exchange: Solving quintic equations with elliptic functions.

[@wernergamper6200]

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