Factoring Sum of Cube

**vaironxxrd** · September 25th 2011, 06:18 AM

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: $5x^3+25x+125$

**Prove It** · September 25th 2011, 06:26 AM

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 $\displaystyle x^3 + 5^3$ ...

**Plato** · September 25th 2011, 06:27 AM

Originally Posted by vaironxxrd

I was given the formula $a^3+b^3=&(a+b)(a^2-ab+b^2)$ For cubes.
Problem: $x^6+125$

$x^6+125=(x^2+5)(x^4-5x^2+25)$

**vaironxxrd** · September 25th 2011, 06:55 AM

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.

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