Answer
Yes if the $x^{3}$ is irrational, then $x$ is irrational.
Work Step by Step
Using proof by contradiction
Property: if $a$ is a rational number then there exist integer $b$ and $c$ such that
$a=\frac{b}{c}$
$x^{3}$ is irrational and thus there do $Not$ exist integers $v$ and $w$ such that
$x^{3}=\frac{v}{w}$
Let us assume that $x$ is rational (if we obtain a contradiction, then this assumption is false and thus then $x$ is irrational). There exists integers $y$ and $ z$, such that:
$x=\frac{y}{z}$
Take the 3rd power of each side of the previous equation:
$x^{3}=\left(\frac{y}{z}\right)^{3}$
as we know that the power of a fraction is the fraction of the power of nominator and the power of the denominator.
$x^{3}=\frac{y^{3}}{z^{3}}$
Since $y$ and $z$ are integers, and $y^{3}$ and $z^{3}$ are also integers and thus we have shown that $x^{3}$ is rational. However $x^{3}$ is known to be irrational and thus we obtained a contradiction.
Hence the assumption that $x$ is rational is then not correct and thus $x$ is irrational.
