Is the cube root of a prime number rational? The question is: if $P$ is prime, is $P^{1/3}$ rational? 
I have been able to prove that if $P$ is prime then the square root of $P$ isn't rational (by contradiction) how would I go about the cube root? 

Suppose $\sqrt[3]{P} = \dfrac{a}{b}$ where $a$ and $b$ have no common factors (i.e. the fraction is in reduced form). Then you have
$$
b^3 P = a^3.
$$
Both sides must be divisible by $a$ (if they're both equal to $a^3$). We already know that $a$ does not divide $b$ (when we assumed the fraction is reduced). So then $a$ must divide $P$.
EDIT: and if $a = 1$, then $P = \dfrac{1}{b^3}$. How many integers are of the form $\dfrac{1}{B}$ for some $B$?
