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!
 A: (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.
A: A slightly cleaner version of the above answers:
Assume $\sqrt[3]{pq} = \frac {a}{b}$ is in reduced form.  Then
$$b^3 p q = a^3.$$
The number of factors of $p$ on the left cannot equal the number of factors of $p$ on the right.  (See why?)
From the Fundamental Theorem of Arithmetic (unique factorization of integers) we have a contradiction.
A: If $x$ is the cube root. Write $x={a\over b}$, $a,b$ relatively prime. We have $x^3=pq$ implies that $a^3=b^3pq$ implies that $p,q$ divides $a$, $a=pqu$ and $a^3=(pq)^3u^3=b^3pq$ implies that $(pq)^2u^3=b^3$ and $p,q$ divides $b$ contradiction.
A: Just do it the same way you'd do square roots. I.e., let $a$ be any number that is $n$-power-free, and consider it's $n$-th root. For the sake of eventual contradiction, assume $a^{1/n}$ is rational, say $a^{1/n} = u / v$ with $\gcd(u, v) = 1$. So:
$\begin{align*}
  a
    &= \frac{u^n}{v^n} \\
  a v^n
    &= u^n
\end{align*}$
Now the left hand side is divisible by $u^n$, thus $a$ is divisible by $u^n$. Contradiction, we assumed $a$ is free of $n$-th powers.
Your $a = p q$ for $p, q$ primes is clearly cube-free.
