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? 
 A: 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 $B$?
A: The main point is: The cube root of a natural number is rational iff it is infact an integer. More generally, any rational root of a monic  polynomial with integer coefficients (such as $X^3-n$) is in fact integer. So if $\sqrt[3] n$ is rational then $n$ is a cube (and cannot be prime).
A: Hint $ $ Any rational root of $\,x^3-p\,$  is an integer, by the  Rational Root Test.
Alternatively $\, a^3 = pb^3\,$ contradicts the uniqueness of prime factorizations, since the prime $\,p\,$ occurs to power a multiple of $\,3\,$ on the lhs, but a nonmultiple $\,1\!+\!3n\,$ on  rhs, i.e. $\,0\not\equiv 1\pmod 3.\,$  This is a generalization of the analogous proof of irrationality of square-roots by comparing the parity of exponents of $\,p,\,$ i.e. $\,0\not\equiv 1\pmod 2,\,$ i.e. even $\ne $ odd. Precisely the same proof works for $k$'th roots, by employing that $\ 0\not\equiv 1\pmod{\! k}$
