How to solve a floor equation?

Question

I am trying to solve the following equation, but I've found some difficulties.

$$\lfloor\sqrt x\rfloor=\lfloor\sqrt[3] x\rfloor\quad\quad(x\in \mathbb R)$$

Answer 1

If $x\lt0$, $\sqrt{x}$ is undefined hence the question makes no sense. For every $x$ in $[0,1)$, $\sqrt{x}$ and $\sqrt[3]{x}$ are both in $[0,1)$ hence $\lfloor\sqrt{x}\rfloor=\lfloor\sqrt[3]{x}\rfloor=0$.

If $x\gt1$, $\lfloor\sqrt{x}\rfloor=\lfloor\sqrt[3]{x}\rfloor$ if and only if there exists an integer $n$ such that $n\leqslant\sqrt[3]{x}\leqslant\sqrt{x}\lt n+1$. Since $x\gt1$, one knows that $n\geqslant1$. The double inequality is equivalent to $n^3\leqslant x\lt(n+1)^2$, in particular one must have $n^3\lt(n+1)^2$. For $n\geqslant1$, this means that $n=1$ or $n=2$. Thus, every $x$ such that $1^3\leqslant x\lt(1+1)^2$ or $2^3\leqslant x\lt(2+1)^2$ is a solution.

Finally, the set of solutions is $[0,4)\cup[8,9)$.