Proof that cube root of product of 2 prime numbers are irrational I am stuck with this problem from my son's homework:
Given $p$ and $q$ are prime numbers, prove that $\sqrt[3]{pq}$ is irrational
Could someone please shed some light? Thanks!

(You don't explicitly state $p \neq q$.  I assume this here.  A similar argument can be made for the $p = q$ case.)
Suppose $\sqrt[3]{pq} = \frac{a}{b}$ is rational in lowest terms (so $a$ and $b$ are integers, $b > 0$, and $\gcd(a,b) = 1$).  Then \begin{align*}
pq &= \frac{a^3}{b^3}  \\
b^3pq &= a^3
\end{align*}
Since $p$ is prime and $p$ divides the left-hand side, $p$ divides the right-hand side.  A prime that divides $a^3$ must be a prime divisor of $a$, so $a = p a_1$ for some integer $a_1$.  We have
\begin{align*}
b^3pq &= (pa_1)^3  \\
b^3pq &= p^3a_1^3  \\
b^3 &= p^2a_1^3  \\
\end{align*}
By a similar argument as used previously, since prime $p$ divides the right-hand side, it divides $b$.  We have $p$ divides $a$ and $p$ divides $b$, so $\gcd(a,b) \geq p > 1$.  This contradicts that $a/b$ is in lowest terms.  Therefore, $\sqrt[3]{ab}$ is irrational.
