Demetres wrote: Let $x,y$ be real numbers with $x \geqslant \sqrt{2021}$ such that \[ \sqrt[3]{x+\sqrt{2021}}+\sqrt[3]{x-\sqrt{2021}} = \sqrt[3]{y}\]Determine the set of all possible values of $y/x$. Let $u=\sqrt[3]{\frac yx}$ Equation $\iff$ (cubing) $y=2x+3\sqrt[3]y\sqrt[3]{x^2-2021}$ (note that this is equivalence because two summands in original LHS can not be equal) $\iff$ $u^3=2+3ut$ where $t=\sqrt[3]{1-\frac{2021}{x^2}}\in[0,1)$ (remember $x>0$) Writing this as $t=g(u)$ with $g(x)=\frac{x^3-2}{3x}$, we have : $\forall u<0$ : $g(u)\ge 1$ and so does not fit $\forall u>0$ : $g(u)$ is continuous increasing and $g(\sqrt[3]2)=0$ and $g(2)=1$ And so $t\in[0,1)$ $\iff$ $u\in[\sqrt[3]2,2)$ and, since $\frac yx=u^3$ : Answer $\boxed{\frac yx\in[2,8)}$