The Play of Roots

$x^3 +4x -1 = 0$ has roots $\alpha , \beta, \gamma$ and $6y^3 -7y^2+ 3y -1 =0$ where $y= \frac {1}{1+x}$ . If $\frac {(\beta+1)(\gamma+1)}{(\alpha+1)^2} + \frac {(\gamma+1)(\alpha+1)}{(\beta+1)^2} + \frac {(\alpha+1)(\beta+1)}{(\gamma+1)^2}$ can be expressed as $\frac {a}{b}$ where a and b are coprime integers, find $a+b$ .