Question: I need to factor:
$$\frac{1}{64}x^3-\frac{3}{8}x^2y+3xy^2-8y^3$$

Tried to do it with an identity but failed,

Factor theorem maybe ?

Thanks

Answer: 
**Hint.** You may try to use the following binomial identity
$$
(a - b)^3=a^3-3 a^2 b+3 a b^2-b^3.
$$
As suggested by Olivier Oloa,

Try $(a-b)^3 = a^3 -3a^2b+3ab^2 -b^3$.

Compare $ a^3 -3a^2b+3ab^2 -b^3$ with $\frac{x^3}{64} - \frac{3x^2y}{8}+3xy^2-8y^3$ 

Thats gives you $ a = \frac{x}{4}$ and $b = 2y$

I'm sure you can take it from there.