# How do you factor 64z − z^3 ?

Jun 15, 2016

$z \left(8 + z\right) \left(8 - z\right)$

#### Explanation:

First of all, you can factor a $z$, to obtain

$z \left(64 - {z}^{2}\right)$

Inside the parenthesis now we have the difference of two squares: we know that ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, and thus

$\left(64 - {z}^{2}\right) = \left({8}^{2} - {z}^{2}\right) = \left(8 + z\right) \left(8 - z\right)$

Since now all three factors $z$, $8 + z$ and $8 - z$ are linear (i.e. polynomials of first degree), we can no longer simplify the answer.