Alpha, Beta Gamma strike again

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

Noting that $x=1$ is not a solution to the equation $x^3-x-1=0$ , it can be 'safely' said that

$y(1-x)=1+x \implies x=\frac{y-1}{y+1}$

Substituting this value of $x$ in the original equation, we get

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

This simplifies to $-\boxed{(y^3+7y^2-y+1)=0}$

From Vieta's formula, we know that

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

Now, we want to form an equation with roots $\dfrac{1+\alpha}{1-\alpha}, \dfrac{1+\beta}{1-\beta},\dfrac{1+\gamma}{1-\gamma}$

To simplify things, we let $\alpha = a+1,\beta=b+1,\gamma = c+1$

Our roots now are:

$\dfrac{1+(a+1)}{1-(a+1)},\dfrac{1+(b+1)}{1-(b+1)},\dfrac{1+(c+1)}{1-(c+1)} = -\dfrac{a+2}{a},-\dfrac{b+2}{b},-\dfrac{c+2}{c}$

The equations above become (do this yourself!):

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

Now, we find the coefficients for the new equation:

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

$\left(-\dfrac{a+2}{a}\right)\left(-\dfrac{b+2}{b}\right)+\left(-\dfrac{a+2}{a}\right)\left(-\dfrac{c+2}{c}\right)+\left(-\dfrac{b+2}{b}\right)\left(-\dfrac{c+2}{c}\right)\\ =\dfrac{(a+2)(b+2)}{ab} + \dfrac{(a+2)(c+2)}{ac} + \dfrac{(b+2)(c+2)}{bc}\\ =\dfrac{c(ab+2a+2b+4)+b(ac+2a+2c+4)+a(bc+2b+2c+4)}{abc}\\ =\dfrac{3abc+4(ab+ac+bc)+4(a+b+c)}{abc}\\ =\dfrac{3(1) + 4(2) + 4(-3)}{1}\\ =-1$

$\left(-\dfrac{a+2}{a}\right)\left(-\dfrac{b+2}{b}\right)\left(-\dfrac{c+2}{c}\right)\\ =-\dfrac{(a+2)(b+2)(c+2)}{abc}\\ =-\dfrac{abc+2ab+2ac+2bc+4a+4b+4c+8}{abc}\\ =-\dfrac{1+2(2)+4(-3)+8}{1}\\ =-1$

Use these values to form the equation:

$x^3 - (-7)x^2 + (-1) x - (-1) = 0\\ \boxed{x^3 + 7x^2 - x + 1 = 0}$