Inspired by Akshat Sharda

First of all, note that $x=1$ is not a root to the equation $x^{3}-x-1=0$ , so, the value of given expression is evaluate-able.

Notice that $\dfrac{(1+\alpha)^{2}}{1-\alpha}\,=\,\dfrac{(1-\alpha)^{2}+4\alpha}{1-\alpha}\,=\,1-\alpha+4\left(\dfrac{\alpha}{1-\alpha}\right)$ . Similarly, $\dfrac{(1+\beta)^{2}}{1-\beta}\,=\,1-\beta+4\left(\dfrac{\beta}{1-\beta}\right)$ and $\dfrac{\gamma^{2}}{1-\gamma}\,=\,1-\gamma+4\left(\dfrac{\gamma}{1-\gamma}\right)$ .

So, $\sum_{\text{cyc}}\dfrac{(1+\alpha)^{2}}{1-\alpha}\,=\,3-(\alpha+\beta+\gamma)+4\left(\dfrac{\alpha}{1-\alpha}+\dfrac{\beta}{1-\beta}+\dfrac{\gamma}{1-\gamma}\right)$ .

Now, replacing $x$ by $\dfrac{x}{1+x}$ in the equation $x^{3}-x+1\,=\,0$ followed by appropriate simplifications gives $x^{3}+5x^{2}+4x+1\,=\,0$ . Clearly, this equation has roots of the form $\dfrac{\alpha}{1-\alpha}$ . Now, Viete's Theorem, $\dfrac{\alpha}{1-\alpha}+\dfrac{\beta}{1-\beta}+\dfrac{\gamma}{1-\gamma}\,=\,-5$ . So, $\sum_{\text{cyc}}\dfrac{(1+\alpha)^{2}}{1-\alpha}\,=\,3-0+4\cdot(-5)\,=\,\boxed{-17}$ .

$\displaystyle \sum_{n=\{\alpha,\beta,\gamma\}} \dfrac{(1+n)^2}{1-n}\\ =\dfrac{(1+\alpha)^2}{1-\alpha}+\dfrac{(1+\beta)^2}{1-\beta}+\dfrac{(1+\gamma)^2}{1-\gamma}$

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:

$\dfrac{(a+2)^2}{-a} + \dfrac{(b+2)^2}{-b} + \dfrac{(c+2)^2}{-c}\\ =-\left(\dfrac{bc(a^2+4a+4) + ac(b^2 + 4b+4) + ab(c^2 + 4c + 4)}{abc}\right)\\ =-\left(\dfrac{a^2bc + ab^2c + abc^2 + 12abc + 4ab + 4ac + 4bc}{abc} \right)\\ =-\left(\dfrac{abc(a+b+c) + 12abc + 4(ab+ac+bc)}{abc}\right)$

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:

$-\left(\dfrac{abc(a+b+c) + 12abc + 4(ab+ac+bc)}{abc}\right)\\ =-\left(\dfrac{1(-3) + 12(1) + 4(2)}{1}\right)\\ =\boxed{-17}$