Gazeta Matematica 3/2017, Problem 27353

Problem:
Solve in real numbers the equation $$\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}=\dfrac{x(x^2+3)}{3x^2+1}.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
 Let $f(x)=\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}$. An easy check shows that $f$ is a one-to-one correspondence on $\mathbb{R}$. Let us prove that $f^{-1}(x)=\dfrac{x(x^2+3)}{3x^2+1}$. Indeed, let us solve with respect to $x$ the equation $y=f(x)$. We have $$\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}} = y.$$ If $y=1$, we have $1-x=0$, i.e. $x=1$. Assume that $y \neq 1$. Set $u=\sqrt[3]{1+x}$ and $v=\sqrt[3]{1-x}$. Hence, $$\dfrac{u-v}{u+v}=y \implies u=\dfrac{v(y+1)}{1-y}.$$ Since $u^3+v^3=2$, then $$v^3\dfrac{(y+1)^3}{(1-y)^3}+v^3=2 \implies v^3=\dfrac{(1-y)^3}{3y^2+1}.$$ Therefore, $1-x=\dfrac{(1-y)^3}{3y^2+1}$, which gives $x=\dfrac{y(y^2+3)}{3y^2+1}$, so $f^{-1}(x)=\dfrac{x(x^2+3)}{3x^2+1}$. It follows that the real solutions to this equation are on the intersection between one of the two functions and the line $g(x)=x$. Let us take $f^{-1}(x)$ for simplicity. Then, it must be $$\dfrac{x(x^2+3)}{3x^2+1}=x.$$ Clearly, $x=0$ is a solution. If $x \neq 0$, then $x^2+3=1+3x^2$, which gives $x=\pm 1$. So, $x \in \{-1,0,1\}$.