= Solution
Set $y=f(x)$ and cube the equation. Solving for $x^3$ gives
$$
y^3=\frac{x^3}{1+x^3},\qquad
x^3=\frac{y^3}{1-y^3}.
$$
Consequently the <inverse function> is
$$
\boxed{f^{-1}(y)=\frac{y}{\sqrt[3]{1-y^3}},\qquad y\ne1.}
$$
The value $y=1$ is impossible in the original equation. Conversely, every real $y\ne1$ gives exactly one real $x$ by the displayed <cube root>, and this $x$ cannot equal $-1$. Direct substitution verifies both inverse identities. Thus the corrected <bijection> is $f:\mathbb R\setminus\{-1\}\to\mathbb R\setminus\{1\}$.
Back to article page
