<s>[INST] 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 $$f(x)=f^{-1}(x)=x$$
 [/INST] $$ \frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $$
Use componendo-dividendo:
$$ \frac{\sqrt[3]{1+x} -\sqrt[3]{1-x} + (\sqrt[3]{1+x} +\sqrt[3]{1-x})}{\sqrt[3]{1+x} -\sqrt[3]{1-x} - (\sqrt[3]{1+x} +\sqrt[3]{1-x})} = \frac{x(x^2+3) + 3x^2+1}{x(x^2+3) -(3x^2+1)} $$
$$ \frac{2\sqrt[3]{1+x}}{-2\sqrt[3]{1-x}} = \frac{x^3+3x^2+3x+1}{x^3-3x^2+3x-1}$$
$$ -\sqrt[3]{\frac{{1+x}}{{1-x}}} = \frac{(x+1)^3}{(x-1)^3} = -\left(\frac{1+x}{1-x}\right)^3$$
I think you can handle it from here
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