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 ?
On
Since the polynomial $x^3-x-1$ is irreducible over$\def\Q{\Bbb Q}~\Q$ by the rational root test, one approach would be to identify the element $\frac{1+a}{1-a}$ where $a$ is the image of $x$ in the field $K=\Q[x]/(x^3-x-1)$, and to compute its minimum polynomial over$~\Q$; since the Galois group of the splitting field of $x^3-x-1$ permutes its roots transitively, that polynomial will also have as roots the elements obtained by substituting another root ($b$ or $c$) for$~a$.
To compute the inverse of $1-a$ in $K$, find the Bézout coefficients of $\gcd(1-x,x^3-x-1)=1$ which are $-x-x^2$ and $-1$ since $1=(-x-x^2)(1-x) -(-1-x+x^3)$; the first one gives the inverse $-a-a^2$ of $1-a$. Now compute $ q=\frac{1+a}{1-a}=(1+a)(-a-a^2)=-a-2a^2-a^3$, which reduces to $q=-1-2a-2a^2$ using $a^3=a+1$. Using the same relation, one finds the square and cube of $q$ to be respectively $q^2=9+16a+12a^2$ and $q^3=-65-114a-86a^2$. Now simple linear algebra finds the linear dependence of $1,q,q^2,q^3$ to be $1-q+7q^2+q^3=0$ so the polynomial that was asked for is $1-x+7x^2+x^3$.
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.