The Play of Roots

The expression equals

$\large \frac { (\alpha + 1)^3(\beta + 1)^3 + (\alpha + 1)^3 (\gamma + 1)^3 + (\alpha + 1)^3 (\alpha + 1)^3 }{ \left ( (\alpha + 1)(\beta + 1) (\gamma + 1) \right )^2}$

$\large = (\alpha + 1)(\beta + 1)(\gamma + 1) \cdot \left ( \frac {1}{(\alpha + 1)^3} + \frac {1}{(\beta + 1)^3} + \frac {1}{(\gamma + 1)^3} \right )$

Let $f(x) = x^3 + 4x - 1$ , then $f(x)$ has roots $\alpha, \beta, \gamma$ .

$\begin{aligned} \Rightarrow f(x-1) & = & (x-1)^3 + 4(x-1) - 1 \\ & = & x^3 -3x^2 +7x - 6 \\ \end{aligned}$

has roots $(\alpha + 1), (\beta + 1), (\gamma + 1)$

By Vieta's formula, $(\alpha + 1)(\beta + 1)(\gamma + 1) = 6$

Let $g(x) = x^3 - 3x^2 + 7x - 6$ , then $g(\frac {1}{x} ) = 0$ has roots $\frac {1}{\alpha + 1},\frac {1}{\beta + 1},\frac {1}{\gamma + 1}$

$\begin{aligned} \Rightarrow \frac {1}{x^3} - \frac {3}{x^2} + \frac {7}{x} - 6 & = & 0 \\ 6x^3 - 7x^2+3x-1 & = & 0 \\ \end{aligned}$

has roots $\frac {1}{\alpha + 1},\frac {1}{\beta + 1},\frac {1}{\gamma + 1}$

Recall the identity, $X^3 + Y^3 + Z^3 - 3XYZ = (X+Y+Z) \left ( (X^2+Y^2+Z^2) - (XY+XZ+YZ) \right )$

In this case, let $X,Y, Z$ be $\frac {1}{\alpha + 1},\frac {1}{\beta + 1},\frac {1}{\gamma + 1}$

$\begin{aligned} \Rightarrow X^3 + Y^3 + Z^3 - 3\left ( \frac {1}{6} \right ) & = & \left ( \frac {7}{6} \right ) \left ( (X+Y+Z)^2 - 3(XY+XZ+YZ) \right ) \\ X^3 + Y^3 + Z^3 - \frac {1}{2} & = & \left ( \frac {7}{6} \right ) \left ( \left ( \frac {7}{6} \right )^2 - 3 \left ( \frac {3}{6} \right ) \right ) \\ X^3 + Y^3 + Z^3 & = & \frac {73}{216} \\ \end{aligned}$

Hence, the expression equals to $6 \cdot \frac {73}{216} = \frac {73}{36}$

The answer is $73 + 36 = \boxed{109}$