February, 2017

Solution:
Let $f(x)=\frac{x^{3}-8}{8}$. Then $f^{-1}(x)=\sqrt[3]{8x+8}=2\sqrt[3]{x+1}$, and so the given equation is equivalent to $f(x)=f^{-1}(x)$. This implies $f(f(x))=x$. However, as $f$ is monotonically increasing, this implies that $f(x)=x$.

As a result, we have
$$\frac{x^{3}-8}{8}=x \Longrightarrow x^{3}-8x-8=0 \Longrightarrow (x+2)\left(x^{2}-2x-4\right)=0,$$
and so $x=-2, 1 \pm \sqrt{5}$.