# 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 ?

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.
• @MarcvanLeeuwen Indeed $(y-1)(y+1)^2=y^3+y^2-y-1$ and not $y^3-1$ as seems to be implied by the penultimate line, and this gives the same as your method. May 2, 2015 at 13:20
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 $$1-x+7x^2+x^3$$.