# Thread: Factoring cubes

1. ## Factoring cubes

Hi guys. I've been wondering if you could possibly help me. I don't know what to do with that 27 on the second term.

2. Think of it as $(4)^3 + (3a)^3$

3. That's really useful...

4. Originally Posted by Alienis Back
That's really useful...
Yes it is.

Maybe I should spell it out for you: you should treat 3a as one. Substitute x = 3a...

$4^3 + x^3 = (4+x)(16-4x-x^2)$

5. Originally Posted by Chop Suey
Yes it is.

Maybe I should spell it out for you: you should treat 3a as one. Substitute x = 3a...[/tex]
That is a good idea.

Originally Posted by Chop Suey
$4^3 + x^3 = (4+x)(16-4x-x^2)$
But I think here there is a mistake. Look at the formula for the sum of cubes above.

6. Originally Posted by Alienis Back
That is a good idea.

But I think here there is a mistake. Look at the formula for the sum of cubes above.
Yes, you are right. I made a typo there. it should be +x^2 rather than -x^2.