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? 

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 $\ 0\not\equiv 1\pmod{\! k}$
