<s>[INST] Is Vieta the only way out? Let $a,b,c$ are the three roots of the equation $x^3-x-1=0$. Then find the equation whose roots are $\frac{1+a}{1-a}$,$\frac{1+b}{1-b}$,$\frac{1+c}{1-c}$.
The only solution I could think of is by using Vieta's formula repeatedly which is no doubt a very messy solution. Is there any easier and more slick way of doing this ?
 [/INST] Transform the equation. Since all the roots are symmetric, say $$y=\frac {1+x}{1-x}\implies x(y+1)=y-1\implies x=\frac {y-1}{y+1}$$ Substitute this expression in place of $x$ in the original equation and simplify. $$\require{cancel}\begin{align}f(y)&=\biggl(\frac {y-1}{y+1}\biggr)^3-\frac {y-1}{y+1}-1=0\\&\implies(y-1)^3-(y+1)^3-(y-1)(y+1)^2=0\\&\implies (y+1)^3-(y-1)^3+(y-1)(y+1)^2=0\\&\implies \cancel{y^3}+3y^2+\bcancel{3y}+1-\cancel{y^3}+3y^2-\bcancel{3y}+\xcancel{1}+y^3+y^2-y-\xcancel{1}=0\\&\implies y^3+7y^2-y+1=0\end{align}$$This is the required answer.
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