Proof that $\sqrt[3]{17}$ is irrational

If you are allowed to use the uniqueness of prime factorisations then an equivalent argument is as follows:

In the prime factorisation of any cube, the exponent of each prime must be a multiple of $3$ i.e. it is $0, 3, 6, 12$ etc. In particular, the exponents of $17$ in the prime factorisations of $p^3$ and $q^3$ must each be a multiple of $3$.

But since $17q^3 = p^3$, the exponents of $17$ in the prime factorisations of $p^3$ and $q^3$ must differ by $1$ (otherwise $p^3$ would have two different prime factorisations). Two multiples of $3$ cannot differ by $1$, so $p$ and $q$ such that $17q^3 = p^3$ do not exist.

Using same idea have used here. It's clear that, $17^{1/3}$ is root of the monic polynomial $x^3-17=0 $. Now, if $17^{1/3}$ is an rational algebraic number, it need to be an integer. But, $2^3=8<17<3^3=27$; so, $2<\sqrt[3]{17}<3$. Hence, it is a irrational number.