hi how do you factor out (x^3-64)

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- Sep 6th 2008, 01:26 PMchicanala2290factoring
hi how do you factor out (x^3-64)

- Sep 6th 2008, 01:30 PMMatt 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 PMKrizalid
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 PM11rdc11
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 PMKrizalid
What about expanding $\displaystyle (x-y)^3?$ That yields the LHS.

- Sep 6th 2008, 02:15 PM11rdc11
Oops lol I see what you mean. That was a dumb question on my part.