Find all possible solutions of $\sqrt[3]{(x^2 + x - 1)} + \sqrt[3]{(10 - x - x^2)} = 3$ The Question: Find all possible solutions $x$ to the equation $\sqrt[3]{(x^2 + x - 1)} + \sqrt[3]{(10 - x - x^2)} = 3$.
My answer:
Let $x^2 + x - 1 = y$.
Then, $\sqrt[3]{y} + \sqrt[3]{9 - y} = 3.$
So, $\sqrt[3]{9 - y} = 3 - \sqrt[3]{y}$.
Cubing both sides, we get, $9 - y = 27 - 27\sqrt[3]{y} + 9(\sqrt[3]{y})^2 - y.$
Simplifying and dividing by $9$, we get $(\sqrt[3]{y})^2 - 3\sqrt[3]{y} + 2 = 0.$
Factoring, we get: $(\sqrt[3]{y} - 1)(\sqrt[3]{y} - 2) = 0.$
Therefore, we get $y = 1, \sqrt[3]{2}$.
Substituting this back into the equation $x^2 + x - 1 = y$, we get $\boxed{x = 1, -2,}$ and $\boxed{{\frac{-1 \pm \sqrt{37}}{2}}}$.

Alternatively, let $x^2+x=t+\frac{11}2$ to get the symmetric equation in $t$
$$\sqrt[3]{\frac92-t} + \sqrt[3]{\frac92+t} = 3 \implies 
\sqrt[3]{\left(\frac{9}2\right)^2-t^2}=2\implies t=\pm \frac72
$$
which leads to $x^2+x=9$ and $x^2+x=2$.
