Effective method to solve $\frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $ I want to know, is there an easy method to solve below equation $$\frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $$ I tried it by plotting and find the solution $x=0,1,-1$ then I tried to solve it by sustitution, multiplying by $(\sqrt[3]{(1+x)^2}+\sqrt[3]{1+x} \sqrt[3]{1-x} +\sqrt[3]{(1-x)^2})$ and so on ...
but I got stuck in solving it normally.
I am thankful if anyone can show me the clue.
Implicit:  when I plot both sides , I feel if we name $$f(x)=\frac{(\sqrt[3]{1+x} -\sqrt[3]{1-x})}{(\sqrt[3]{1+x} +\sqrt[3]{1-x})}$$ then  $$f^{-1}= \frac{x(x^2+3)}{3x^2+1} $$  so It is an effective way to solve $$f(x)=f^{-1}(x)=x$$

I think about it again , $$f(x)=\frac{(\sqrt[3]{1+x} -\sqrt[3]{1-x})}{(\sqrt[3]{1+x} +\sqrt[3]{1-x})}=\\\frac{(\sqrt[3]{\frac{1+x}{1-x}} -1)}{(\sqrt[3]\frac{1+x}{1-x} +1)}$$so $$y=\frac{(\sqrt[3]{\frac{1+x}{1-x}} -1)}{(\sqrt[3]\frac{1+x}{1-x} +1)}=\frac{a+1}{a-1}\\\to a=\frac{y+1}{y-1}\\\to  \sqrt[3]{\frac{1+x}{1-x}}=\frac{y+1}{y-1}\\\to x=y\\x=\frac{x(x^2+3)}{3x^2+1}
\\ \to x=0,1,-1$$
