François Viete Going Wild

Let $y =\dfrac{1+\alpha}{1-\alpha}\implies \alpha=\dfrac{y-1}{y+1}$

Since $\alpha$ is a root of the original equation, $\therefore \left(\dfrac{y-1}{y+1}\right)^3-\left( \dfrac{y-1}{y+1}\right)-1=0$

$\implies y^3+7y^2-y+1=0$

The sum of roots of this equation is $-7$ which is the required value. Hence:-

$\dfrac{1\ + \ \alpha}{1 \ - \ \alpha} + \dfrac{1 \ + \ \beta}{1\ - \ \beta} + \dfrac{1 \ +\ \gamma}{1 \ - \ \gamma}=\boxed{- 7}$

See Transforming Roots of a polynomial

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$\LARGE\star\color{#69047E}{\boxed{\mathscr{\color{#007fff}{Alternate~Solution}}}}\star$ Call the expression $\mathfrak F$

Taking LCM and slight simplification gives:-

$\mathfrak{F}=\dfrac{3+3\alpha\beta\gamma-2\displaystyle\sum_{\text{cyc}}\alpha-\displaystyle\sum_{\text{cyc}}\alpha\beta}{1+\displaystyle\sum_{\text{cyc}}\alpha\beta-\displaystyle\sum_{\text{cyc}}\alpha-\alpha\beta\gamma}$

Using Vietas, $\displaystyle\sum_{\text{cyc}}\alpha=0,\displaystyle\sum_{\text{cyc}}\alpha\beta=-1,\alpha\beta\gamma=1$ .

Substituting we get $\large \mathfrak{F}=\boxed{- 7}$ .

(See Vieta's Formula )

Source of this solution

$\displaystyle \sum_{n=\{\alpha,\beta,\gamma\}} \dfrac{1+n}{1-n}\\ =\dfrac{1+\alpha}{1-\alpha}+\dfrac{1+\beta}{1-\beta}+\dfrac{1+\gamma}{1-\gamma}$

From Vieta's formula, we know that

$\alpha + \beta + \gamma = 0\\ \alpha\beta+\alpha\gamma + \beta\gamma = -1\\ \alpha\beta\gamma = 1$

Now, to simplify our calculations, we substitute $\alpha = a + 1$ , $\beta = b+1$ and $\gamma = c+1$ into the expression and the equations

The expression becomes:

$\dfrac{a+2}{-a} + \dfrac{b+2}{-b} + \dfrac{c+2}{-c}\\ =-\left(\dfrac{bc(a+2) + ac(b+2) + ab(c+2)}{abc}\right)\\ =-\left(\dfrac{3abc+2(ab+ac+bc)}{abc} \right)$

The equations become (simplify yourself, I'm too lazy to type out all the steps):

$a + 1 + b + 1 + c + 1 = 0 \implies a+b+c = -3\\ (a+1)(b+1) + (a+1)(c+1) + (b+1)(c+1) = -1 \implies ab+ac+bc = 2\\ (a+1)(b+1)(c+1) = 1 \implies abc=1$

Substitute these values into the expression to get the answer:

$-\left(\dfrac{3abc+2(ab+ac+bc)}{abc} \right)\\ =-\left(\dfrac{3(1)+2(2)}{1}\right)\\ =\boxed{-7}$

We can make a equation having its roots as $\frac{1+\alpha}{1-\alpha},\frac{1+\beta}{1-\beta}$ and $\frac{1+\gamma}{1-\gamma}$ .

Let $y=\frac{1+x}{1-x}$ , then $x=\frac{y-1}{y+1}$ .

Placing $x=\frac{y-1}{y+1}$ in $x^3-x-1=0$ .

$\left(\frac{y-1}{y+1}\right)^3-\left(\frac{y-1}{y+1}\right)+1=0 \\ (y-1)^3-(y-1)(y+1)^2-(y+1)^3=0 \\ y^3+7y^2-y+1=0$

So, $y^3+7y^2-y+1=0$ has $\frac{1+\alpha}{1-\alpha},\frac{1+\beta}{1-\beta}$ and $\frac{1+\gamma}{1-\gamma}$ as its roots.

Sum of roots of this equation $=\boxed{-7}$

$\therefore \dfrac{1 + \alpha}{1 - \alpha} + \dfrac{1 + \beta}{1 - \beta} + \dfrac{1 + \gamma}{1 - \gamma}=\boxed{-7}$ .

$S\quad =\quad \frac { 1+\alpha }{ 1-\alpha } \quad +\quad \frac { 1+\beta }{ 1-\beta } \quad +\quad \frac { 1+\gamma }{ 1-\gamma } \\ T\quad =\quad \frac { 1 }{ 1-\alpha } \quad +\quad \frac { 1 }{ 1-\beta } \quad +\quad \frac { 1 }{ 1-\gamma } \\ First\quad find\quad T\\ f(x)\quad =\quad { x }^{ 3 }-x-1\quad =\quad (x-\alpha )(x-\beta )(x-\gamma )\\ \frac { f'(x) }{ f(x) } \quad =\quad \frac { 3{ x }^{ 2 }-1 }{ { x }^{ 3 }-x-1 } \quad =\quad \frac { 1 }{ x-\alpha } \quad +\quad \frac { 1 }{ x-\beta } \quad +\quad \frac { 1 }{ x-\gamma } \\ So\quad \frac { f'(1) }{ f(1) } \quad =\quad \frac { 2 }{ -1 } \quad =\quad \frac { 1 }{ 1-\alpha } \quad +\quad \frac { 1 }{ 1-\beta } \quad +\quad \frac { 1 }{ 1-\gamma } \quad =\quad T\\ Therefore\quad T=-2\\ S\quad =\quad -(\frac { -1-\alpha }{ 1-\alpha } \quad +\quad \frac { -1-\beta }{ 1-\beta } \quad +\quad \frac { -1-\gamma }{ 1-\gamma } )\quad =\quad \\ =\quad -(\frac { 1-\alpha -2 }{ 1-\alpha } \quad +\quad \frac { 1-\beta -2 }{ 1-\beta } \quad +\quad \frac { 1-\gamma -2 }{ 1-\gamma } )\quad =\\ =\quad -(1\quad +\quad \frac { -2 }{ 1-\alpha } \quad +\quad 1\quad +\quad \frac { -2 }{ 1-\beta } \quad +\quad 1\quad +\quad \frac { -2 }{ 1-\gamma } )\quad =\quad \\ =\quad -(3-2T)\quad =\quad -7$

$\begin{aligned} \text{Rewrite the polynomial,}& \\ x^3 - x - 1 &= ax^3 + bx^3 + cx + d\\ a = 1, \ b & = 0, \ c = -1,\ d=-1 \\ \text{Form the expression into one fraction,}\\ \dfrac{1 + \alpha}{1 - \alpha} + \dfrac{1 + \beta}{1 - \beta} + \dfrac{1 + \gamma}{1 - \gamma} & = \frac{-3 \alpha \beta \gamma + \alpha \beta + \beta \gamma + \alpha \gamma + \alpha + \beta + \gamma -3 }{\alpha \beta \gamma - ( \alpha \beta + \beta \gamma + \alpha \gamma ) + \alpha + \beta + \gamma - 1 } \\ \text{Now look closely and you can see that you can use Vieta's Formula,}\\ \frac{-3 \color{#D61F06}{\alpha \beta \gamma} + \color{#3D99F6}{\alpha \beta + \beta \gamma + \alpha \gamma} + \color{#20A900}{\alpha + \beta + \gamma} -3 }{\color{#D61F06}{\alpha \beta \gamma} - ( \color{#3D99F6}{ \alpha \beta + \beta \gamma + \alpha \gamma }) + \color{#20A900}{\alpha + \beta + \gamma} - 1 } & = \frac{-3 ( \color{#D61F06}{- \frac{d}{a}} ) + ( \color{#3D99F6}{\frac{c}{a}} ) + (\color{#20A900}{- \frac{b}{a}}) -3 }{ (\color{#D61F06}{- \frac{d}{a}}) - (\color{#3D99F6}{\frac{c}{a}}) + (\color{#20A900}{- \frac{b}{a}} ) - 1 }\\ & = \frac{-3 ( \color{#D61F06}{1} ) + ( \color{#3D99F6}{-1} ) + (\color{#20A900}{0}) -3 }{ (\color{#D61F06}{1}) - (\color{#3D99F6}{-1}) + (\color{#20A900}{0} ) - 1 }\\ & = \color{#69047E}{\boxed{-7}} \end{aligned}$

Same as the alternate solution.......