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} $

Question

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 $$f(x)=f^{-1}(x)=x$$

Answer 1

$$ \frac{\sqrt[3]{1+x} -\sqrt[3]{1-x}}{\sqrt[3]{1+x} +\sqrt[3]{1-x}} = \frac{x(x^2+3)}{3x^2+1} $$ 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

Answer 2

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 $$y=\frac{(\sqrt[3]{\frac{1+x}{1-x}} -1)}{(\sqrt[3]\frac{1+x}{1-x} +1)}=\frac{a+1}{a-1}\\\to a=\frac{y+1}{y-1}\\\to \sqrt[3]{\frac{1+x}{1-x}}=\frac{y+1}{y-1}\\\to x=y\\x=\frac{x(x^2+3)}{3x^2+1} \\ \to x=0,1,-1$$