Getting caught up on this problem. what i got so far.
Contrapositive: For $x \in \mathbb{R}$, if $\sqrt[3]{x}$ is rational, then $x$ is rational.
If $\sqrt[3]{x}$ is rational then there must exist an $a,b \in Z, b \neq 0$ such that
$$\sqrt[3]{x}=\frac{a}{b}$$ $$x=\frac{a^3}{b^3}$$ $$b^3\cdot x=a^3$$
As $b^3$ is a multiple of $a^3$, $x$ by defintion can not be rational as they share a common factor.
I think im taking the wrong approach to this problem and should try maybe going by contradiction? or both. Or is the original statement false?
you were already finished at $x=\frac{a^3}{b^3}$, any number of this form is immediately rational.