How do you factor #y^6 - 64z^8#?
To solve this problem you must think what can i factor for the y^6
Since the rule of exponent is y^6
(y)^3 * (y^3) add them
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.