# How do you factor y^6 - 64z^8?

Feb 6, 2016

$\left(8 \cdot {z}^{4} + {y}^{3}\right) \cdot \left(- 8 \cdot {z}^{4} + {y}^{3}\right)$

#### Explanation:

To solve this problem you must think what can i factor for the y^6

Since the rule of exponent is y^6

so y^6

Think of it as y^2
(y) * (y) = y^2
To get 64
what is it divisible by?
Check if it is divisible by 2 to get perfect squares if not
2 * 2 = 4 until 8*8 = 64
So you got the second one.
The factor of z can be the same as above
which is (z^4) (z^4) and then you add them to the parenthesis.
In any order.