# Thread: Factoring Sum of Cube

1. ## Factoring Sum of Cube

I was given the formula $a^3+b^3=&(a+b)(a^2-ab+b^2)$ For cubes.

But this problem is kind of weird for me since I don't know if I should find the cube of $x^6$

Problem: $x^6+125$
Attempt: $(x+5)(x^2+5x+25)$ Distributed: $5x^3+25x+125$

2. ## Re: Factoring Sum of Cube

Originally Posted by vaironxxrd
I was given the formula $a^3+b^3=&(a+b)(a^2-ab+b^2)$ For cubes.

But this problem is kind of weird for me since I don't know if I should find the cube of $x^6$

Problem: $x^6+125$
Attempt: $(x+5)(x^2+5x+25)$ Distributed: $5x^3+25x+125$
It's because $\displaystyle x^6 + 125 = \left(x^2\right)^3 + 5^3$, not $\displaystyle x^3 + 5^3$...

3. ## Re: Factoring Sum of Cube

Originally Posted by vaironxxrd
I was given the formula $a^3+b^3=&(a+b)(a^2-ab+b^2)$ For cubes.
Problem: $x^6+125$
$x^6+125=(x^2+5)(x^4-5x^2+25)$

4. ## Re: Factoring Sum of Cube

Originally Posted by Prove It
It's because $\displaystyle x^6 + 125 = \left(x^2\right)^3 + 5^3$, not $\displaystyle x^3 + 5^3$...
Got it I saw that it's also a way to do it but I really Din't know I could use it here.