2016/17 AHSMC Part 1 Problem 14

The function $f(t) = \sqrt[3]{x^2 + x + t}$, $t \in \mathbb{R}$ is increasing hence $f(f(t)) = t \iff f(t) = t$ that leads to the conclusion that the real solutions of the given equation and $\sqrt[3]{x^2 + x + 3} = 3$ are the same. The solutions of the last equation are just the solutions of the quadratic equation $x^2 + x - 24 = 0$, for which the product of the solutions is $-24$.

Alternative Solution: Let $y = \sqrt[3]{x^2 + x + 3}$ then the given equation can be written as $$y^3 - 3 + y = 27 \iff (y-3)(y^2 + 3y + 10) = 0 \iff y = 3$$ and then $x^2 + x - 24 = 0$. The answer is (B).