# If $α, β, γ$ are roots of $x^3 - x -1 = 0$, then find the value of $\frac{1+α}{1-α} + \frac{1+β}{1-β} + \frac{1+γ}{1-γ}$.

If $α, β, γ$ are roots of $x^3 - x -1 = 0$, then find the value of $$\frac{1+α}{1-α} + \frac{1+β}{1-β} + \frac{1+γ}{1-γ}$$ I found this question asked in a previous year competitive examination, which was multiple choice in nature, the available options to the question were:

1. $1$
2. $0$
3. $-7$
4. $-5$

Considering the time available for a question to be solved in such an examination, is there a way to solve this problem without actually having to expand the the given relation by cross-multiplying the numerators and denominators or even finding the zeroes of the given equation.

• This works the same way as here. – Dietrich Burde Jun 28 '18 at 16:18
• @DietrichBurde The correct answer was given as -7 and not -5. It could verified when done the long way. There's a mistake in the given solution. – Tony1970 Jun 28 '18 at 16:26
• The given solution refers to $x^3-x^2-1$, and not $x^3-x-1$. But the method to solve is the same. Just do it! – Dietrich Burde Jun 28 '18 at 16:28
• Self-contained hint: if $\alpha$ is a root of $x^3-x-1$, find a (cubic) polynomial that $(1+\alpha)/(1-\alpha)$ is a root of. (This is pretty straightforward; if $\alpha'=(1+\alpha)/(1-\alpha)$ then just invert to find $\alpha$ in terms of $\alpha'$, plug that in to the defining polynomial, and collect some terms - it's a little bit of an algebraic slog, but not too bad since it's only a single variable.) Then $\alpha'$, $\beta'$, and $\gamma'$ are roots of this polynomial, and if you have the three roots of a polynomial you should know how to find their sum... – Steven Stadnicki Jun 28 '18 at 16:30
• @DietrichBurde Sorry I didn't immediately notice. But thanks. Could this be done in a yet more simpler way? – Tony1970 Jun 28 '18 at 16:32

## 5 Answers

It is practical to recall that $z\mapsto\frac{1-z}{1+z}$ is an involution. In particular $$\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}$$ is the sum of the reciprocal of the roots of $p\left(\frac{1-x}{1+x}\right)=-\frac{x^3-x^2+7x+1}{(1+x)^3}$, which is also the sum of the reciprocal of the roots of $x^3-x^2+7x+1$. By Vieta's formulas, this is $\color{red}{-7}$ (option 3.).

Hint If $\zeta$ is a root of $p(x)$, then $\zeta' := 1 - \zeta$ is a root of $-p(1 - x)$. Taking $p(x) := x^3 - x - 1$ gives $$q(x) := -p(1 - x) = x^3 - 3 x^2 + 2 x + 1 .$$

Now, $$\frac{1 + \zeta}{1 - \zeta} = \frac{1 + (1 - \zeta')}{\zeta'} = \frac{2}{\zeta'} - 1,$$ so $$S := \frac{1 + \alpha}{1 - \alpha} + \frac{1 + \beta}{1 - \beta} + \frac{1 + \gamma}{1 - \gamma} = 2 \left(\frac{1}{\alpha'} + \frac{1}{\beta'} + \frac{1}{\gamma'}\right) - 3,$$ where we define $\alpha'$ etc. analogously to $\zeta'$ above, so that $\alpha', \beta', \gamma'$ are the roots of $q$.

Now, similarly transform $q(x)$ to find a polynomial $r(x)$ whose roots are $\alpha'' := \frac{1}{\alpha'}$, etc., and observe that $$r(x) = (x - \alpha'')(x - \beta'')(x - \gamma') = x^3 - (\alpha'' + \beta'' + \gamma'')x^2 + \cdots$$ and that the sum is the above expression for the desired quantity $S$ is exactly the negative $\alpha'' + \beta'' + \gamma''$ of the coefficient of the quadratic term.

Additional hint If $\alpha'' := \frac{1}{\alpha'}, \beta'' := \frac{1}{\beta'}, \gamma'' := \frac{1}{\gamma'}$ are the roots of $$r(x) := x^3 q\left(\tfrac{1}{x}\right) = x^3 + 2 x^2 - 3 x + 1,$$ then $\alpha'' + \beta'' + \gamma'' = -2$, and so $$S = 2 (\alpha'' + \beta'' + \gamma'') - 3 = 2(-2) - 3 = -7.$$

Note that $x^3-x-1=0$ is equivalent to $x^3-x=1$ and $x^3-1=x$. That is, for $x\in\{\alpha,\beta,\gamma\}$, we get $$-x-x^2=\frac{x^3-x}{1-x}=\frac{1}{1-x}\text{ and }-1-x-x^2=\frac{x^3-1}{1-x}=\frac{x}{1-x}\,.$$ Consequently, $$\frac{1+x}{1-x}=-1-2x-2x^2\text{ for }x\in\{\alpha,\beta,\gamma\}\,.$$ This means $$\sum_{x\in\{\alpha,\beta,\gamma\}}\,\frac{1+x}{1-x}=-3-2s-2q\,,$$ where $s:=\alpha+\beta+\gamma$ and $q:=\alpha^2+\beta^2+\gamma^2$. It is easy to find $s$ and $q$.

We have $s=0$ and $q=2$.

• Could you elaborate the answer a little bit. I couldn't catch up certain steps you did. – Tony1970 Jun 29 '18 at 1:11
• @Tony1970 You have to point out what you don't understand. I don't know where to start. – Batominovski Jun 29 '18 at 8:28
• How does $- x - x^2 = \frac{x^3 - x}{1-x} = \frac{1}{1-x}$ and why is "and - 1" used. And I did not understand the equivalence of $x^3-x-1=0$ to $x^3-x=1$ and $x^3-1=0$. – Tony1970 Jun 29 '18 at 14:40

Alt. hint:   Let $y = \dfrac{1+x}{1-x} \iff x = \dfrac{y-1}{y+1}$ then:

\begin{align} 0 = (y+1)^3 \cdot P\left(\dfrac{y-1}{y+1}\right) &= (y-1)^3-(y-1)(y+1)^3-(y+1)^3 \\ &= -y^3 - 7 y^2 + y - 1 \end{align}

It follows from Vieta's relations that $\;y_1+y_2+y_3=\ldots$

Since $$\alpha+\beta+\gamma=0,$$ $$\alpha\beta+\alpha\gamma+\beta\gamma=-1$$ and $$\alpha\beta\gamma=1,$$ we obtain: $$\sum_{cyc}\frac{1+\alpha}{1-\alpha}=\frac{\sum\limits_{cyc}(1+\alpha)(1-\beta)(1-\gamma)}{\prod\limits_{cyc}(1-\alpha)}=\frac{\sum\limits_{cyc}(1+\alpha)(1-\beta-\gamma+\beta\gamma)}{1-(\alpha+\beta+\gamma)+(\alpha\beta+\alpha\gamma+\beta\gamma)-\alpha\beta\gamma}=$$ $$=\frac{\sum\limits_{cyc}(1-2\alpha+\alpha\beta+\alpha-2\alpha\beta+\alpha\beta\gamma)}{1-0-1-1}=\frac{\sum\limits_{cyc}(1-\alpha-\alpha\beta+\alpha\beta\gamma)}{-1}=\frac{3+1+3}{-1}=-7.$$