# factoring

• Sep 6th 2008, 01:26 PM
chicanala2290
factoring
hi how do you factor out (x^3-64)
• Sep 6th 2008, 01:30 PM
Matt Westwood
$\displaystyle x^3 - 64 = x^3 - 4^3$

It's a standard equivalence: $\displaystyle x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ or something like that (it's from memory, might not be quite correct, multiply it out to check)

take it away maestro
• Sep 6th 2008, 01:44 PM
Krizalid
Since it's the standard form $\displaystyle x^3-y^3$ and you forgot (suppose) how to factor that, proceed as follows:

\displaystyle \begin{aligned} x^{3}-y^{3}&=(x-y)^{3}+3xy(x-y) \\ & =(x-y)\left\{ \left( x-y \right)^{2}+3xy \right\} \\ & =(x-y)\left( x^{2}-2xy+y^{2}+3xy \right) \\ & =(x-y)\left( x^{2}+xy+y^{2} \right). \end{aligned}

You can also use the polynomial long division but I think this method is much nicer. :D
• Sep 6th 2008, 02:05 PM
11rdc11
I understand how you did everything else but how did you get this

$\displaystyle x^{3}-y^{3}=(x-y)^{3}+3xy(x-y)$

I'm probably just missing something obvious but just can't figure it out.
• Sep 6th 2008, 02:07 PM
Krizalid
What about expanding $\displaystyle (x-y)^3?$ That yields the LHS.
• Sep 6th 2008, 02:15 PM
11rdc11
Oops lol I see what you mean. That was a dumb question on my part.