# Thread: [SOLVED] Simplifying perfect squares

1. ## [SOLVED] Simplifying perfect squares

I know for example (x^2-16) = (x+4) (x-4) but what about something like (x^3-64)? I can't find the rule to solve something like this. I think it involves 3 terms but I can't figure it out.

2. Originally Posted by thekrown
I know for example (x^2-16) = (x+4) (x-4) but what about something like (x^3-64)? I can't find the rule to solve something like this. I think it involves 3 terms but I can't figure it out.
Hi,

For $\displaystyle a^3 \pm b^3$

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

This is known as the sum and difference of two cubes.

In your case, a = x and b = 4

3. (x^3-64)

I keep getting the wrong results.

(x-4)(x^2 -4x + 16) = x^3 - 8x^2 -64
(x+4)(x^2 -4x+16) = x^3 + 64)

4. ## Comment

Be grateful that the problem is as simple as $\displaystyle x^3 = 64$ (meaning no terms that are of the second or first degree in this equation).

You can't simplify this type of problem further and you'll have to know from memory or a table that $\displaystyle 4^3 = 64$.