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[@fernandoderoque1920]

a-b=2
(a-b)^3=8
(a^3-b^3)-3ab(a-b)=8
(98)-3ab(2)=8
6ab=90
ab=15
(ab)^3=15^3
(4x+49)(4x-49)=15^3
16x^2=15^3+49^2
16x^2=5776
x^2=361
x={19,-19}

[@superacademy247]

Wow 😳. Awesome 👌

[@satyendrapuri4758]

a^3-b^3=98 .....(1)
a-b=2 ..............(2)
Dividing (2) by (1)
a^2+ab+b^2=49 ...(3)
a=b+2.
Put this value of 'b' in (3)
(b+2)^3+(b+2)b+b^2=49
=> (b^2±4b+4')(±b^2b)+b^2=49
=> 3b^2+6b+4=49
=>3b^2+6b-45=0
≈> b^2+2b-15=0
Solving quadratic eqn.
b=3 or b= -5 &
a=5 or a= '3

[@superacademy247]

Wow! It looks stunning 👌 quick and precise

[@michaeldoerr5810]

I think that the first comment is better.

[@fernandoderoque1920]

It would be “dividing (1) by (2)”.

[@ritwikgupta3655]

ab=15 can be directly solved for x^2 as ab=(16x^2-49^2)^1/3 and it gives x^2=361, ie x=+/-19. There is no need to solve for a and b.

[@superacademy247]

Wow! So mesmerizing 🙏🙏🙏

[@michaeldoerr5810]

I am going to use that for practice!!! What are the slightly different problems that can be worked out in that way???

[@alephnull8233]

-19 isn't a solution. In general, whenever you take roots while solving an equation you should plug your solution back into the original equation to make sure it holds. In this case, -19 doesn't satisfy the original equation and is considered an "extraneous solution".
Very good video otherwise.

[@superacademy247]

Check again, -19 is one of the solutions. Thanks 🙏🙏🙏

[@alephnull8233]

@@superacademy247 You're right. I got lazy and plugged -19 into the equation using Wolfram Alpha. Gotta be careful when using CAS's.

[@mouradbelkas598]

your B^3 = 4x - 49, not 4x +49, so 4x = -27 +49 , 4x = 22, x = 5.5

[@superacademy247]

Nope

[@klauskarrtofeln6919]

what is the difference between b and p

[@superacademy247]

English is NOT my first language