Thread: Factoring the sum of cubes

1. Factoring the sum of cubes

Hi MHF,

Usually I can factor, but this particular problem is just killing me today. I have been trying to figure this out for over an hour...

$\displaystyle 128r^3+686z^3$

I know two things...

I know I can immediatly factor out the 2 and take it down to $\displaystyle 2(64r^3+343z^3)$.

And I know the answer is going to be in the form of $\displaystyle 2(binomial)(trinomial).$

But for the life of me I cannot seem to find the proper middle terms to make it work.

Thanks in advance.

2. consider $\displaystyle 4^3=64$ and $\displaystyle 7^3 = 343$

now make your problem into the form

$\displaystyle 2(a^3+b^3) = 2(a+b)(a^2-ab+b^2)$

3. Originally Posted by pickslides
consider $\displaystyle 4^3=64$ and $\displaystyle 7^3 = 343$

now make your problem into the form

$\displaystyle 2(a^3+b^3) = 2(a+b)(a^2-ab+b^2)$
$\displaystyle 2(4r+7z)(16r^2-28rz+49z^2)$

Yeah, that should have been obvious. After doing 3 hours of math, even the obvious things become so complicated.

I think it's time to break and brew some coffee! Thank you Pickslides.