7 The roots of the equation \(x ^ { 3 } + 4 x - 1 = 0\) are \(\alpha , \beta\) and \(\gamma\). Use the substitution \(y = \frac { 1 } { 1 + x }\) to show that the equation \(6 y ^ { 3 } - 7 y ^ { 2 } + 3 y - 1 = 0\) has roots \(\frac { 1 } { \alpha + 1 } , \frac { 1 } { \beta + 1 }\) and \(\frac { 1 } { \gamma + 1 }\). For the cases \(n = 1\) and \(n = 2\), find the value of $$\frac { 1 } { ( \alpha + 1 ) ^ { n } } + \frac { 1 } { ( \beta + 1 ) ^ { n } } + \frac { 1 } { ( \gamma + 1 ) ^ { n } }$$ Deduce the value of \(\frac { 1 } { ( \alpha + 1 ) ^ { 3 } } + \frac { 1 } { ( \beta + 1 ) ^ { 3 } } + \frac { 1 } { ( \gamma + 1 ) ^ { 3 } }\). Hence show that \(\frac { ( \beta + 1 ) ( \gamma + 1 ) } { ( \alpha + 1 ) ^ { 2 } } + \frac { ( \gamma + 1 ) ( \alpha + 1 ) } { ( \beta + 1 ) ^ { 2 } } + \frac { ( \alpha + 1 ) ( \beta + 1 ) } { ( \gamma + 1 ) ^ { 2 } } = \frac { 73 } { 36 }\).