# Factoring Completely

• Jul 31st 2009, 05:28 AM
croc870
Factoring Completely
Hello Everyone,

I'm returning to school after a long time off, and I know this is a pretty basic question, but I can't seem to get it.

The problem is pretty simple:

Factor Completely
125-27u^3

Now I know that the cube of 125 is 5 and the cube of 27 is 3, and I'm sure that ties into here since it sounds like I need to find the GCF twice and make three parentheticals. I can't seem to make this work however, and I'm feeling foolish.

Many Thanks!
• Jul 31st 2009, 05:33 AM
yeongil
Actually, this is a difference of cubes, so you use the difference of cubes formula:
$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

So in
$125-27u^3$,

let a = 5 and b = 3u, so
$125-27u^3$
$= 5^3 - (3u)^3$
$= (5 - 3u)(5^2 + 5(3u) + (3u)^2)$
$= (5 - 3u)(25 + 15u + 9u^2)$

01
• Jul 31st 2009, 05:34 AM
stapel
Quote:

Originally Posted by croc870
Factor Completely
125-27u^3

Now I know that the cube of 125 is 5 and the cube of 27 is 3...

I think you mean that the cube of 5, 5^3, is 125 and that the cube of 3, 3^3, is 27...? (Wondering)

Quote:

Originally Posted by croc870
I'm sure that ties into here since it sounds like I need to find the GCF twice and make three parentheticals.

You only need to apply the difference-of-cubes factoring formula to what is, as you have noted, equivalent to (5)^3 - (3u)^3.

You should end up with two sets of parentheses.... (Wink)
• Jul 31st 2009, 05:35 AM
Wilmer
sum difference cubes