Answer
(a) $f^{-1}(x) = \sqrt[3] {\frac{1}{8}x - \frac{1}{8}}$
(b) Both $f(f^{-1}(x))$ and $f^{-1}(f(x))$ result in $x$.
Work Step by Step
$$f(x) = 8x^3 + 1$$ $$y = 8x^3 + 1$$ $$x = 8y^3 + 1$$ $$x-1 = 8y^3$$ $$\frac{1}{8}x - \frac{1}{8} = y^3$$ $$\sqrt[3] {\frac{1}{8}x - \frac{1}{8}} = y$$ $$\frac{1}{8}x - \frac{1}{8} = y^3$$ $$\sqrt[3] {\frac{1}{8}x - \frac{1}{8}} = f^{-1}(x)$$
(b)
$f(f^{-1})(x)$ $$8(\sqrt[3] {\frac{1}{8}x - \frac{1}{8}})^3 + 1$$ $$8(\frac{1}{8}x - \frac{1}{8}) + 1$$ $$x - 1 + 1$$ $$x$$
$f^{-1}(f(x))$ $$\sqrt[3] {\frac{1}{8}(8x^3 + 1) - \frac{1}{8}}$$ $$\sqrt[3] {x^3 + \frac{1}{8} - \frac{1}{8}}$$ $$\sqrt[3] {x^3}$$ $$x$$
