If $\alpha,\beta,\gamma$ are the roots of the equation $x^3-x-1=0$, then the value of $\sum \left(\dfrac {1+\alpha}{1-\alpha}\right)$ is-

Given x^3 - x - 1 = 0, the roots are alpha, beta, gamma. We need the sum of (1+r)/(1-r) for each root. Let y = (1+x)/(1-x), then x = (y-1)/(y+1). Substituting this into the original equation: ((y-1)/(y+1))^3 - (y-1)/(y+1) - 1 = 0. Expanding this leads to a cubic in y, and the sum of roots is -coefficient of y^2 / coefficient of y^3, which results in -7.