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 \(\frac { ( \beta + 1 ) ( \gamma + 1 ) } { ( \alpha + 1 ) ^ { 2 } } + \frac { ( \gamma + 1 ) ( \alpha + 1 ) } { ( \beta + 1 ) ^ { 2 } } + \frac { ( \alpha + 1 ) ( \beta + 1 ) } { ( \gamma + 1 ) ^ { 2 } } = \frac { 73 } { 36 }\).
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\(y = \frac{1}{x+1}\) \(\therefore x = \frac{1-y}{y}\) M1
use in given cubic equation
\(\text{Gives } 6y^3 - 7y^2 + 3y - 1 = 0\) A1
\(n = 1: \text{given expression} = \text{sum of roots} = \frac{7}{6}\) B1
\(n = 2: \sum\frac{1}{(\alpha+1)^2} = \left[\sum\frac{1}{(\alpha+1)}\right]^2 - 2\sum\alpha\beta = \frac{13}{36}\) B1
From cubic in \(y\):
Answer Marks
Guidance
\(6\sum\left(\frac{1}{\alpha+1}\right)^3 - \pi\frac{13}{36} + 3\left(\frac{7}{6}\right) - 3 = 0\) M1
\(\sum\left(\frac{1}{\alpha+1}\right)^3 = \frac{73}{216}\) A1
\(\text{LHS} = \sum\left(\frac{(\beta+1)(\gamma+1)(\alpha+1)}{(\alpha+1)^3}\right)\) M1
\(= \left(\frac{1}{6}\right)^{-1} \times \frac{73}{216}\) M1
recognise product of roots
\(= \frac{73}{36}\) A1
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$y = \frac{1}{x+1}$ $\therefore x = \frac{1-y}{y}$ | M1 | use in given cubic equation
$\text{Gives } 6y^3 - 7y^2 + 3y - 1 = 0$ | A1 | | [2]
$n = 1: \text{given expression} = \text{sum of roots} = \frac{7}{6}$ | B1 | | [2]
$n = 2: \sum\frac{1}{(\alpha+1)^2} = \left[\sum\frac{1}{(\alpha+1)}\right]^2 - 2\sum\alpha\beta = \frac{13}{36}$ | B1 | | [2]
**From cubic in $y$:**
$6\sum\left(\frac{1}{\alpha+1}\right)^3 - \pi\frac{13}{36} + 3\left(\frac{7}{6}\right) - 3 = 0$ | M1
$\sum\left(\frac{1}{\alpha+1}\right)^3 = \frac{73}{216}$ | A1 | | [2]
$\text{LHS} = \sum\left(\frac{(\beta+1)(\gamma+1)(\alpha+1)}{(\alpha+1)^3}\right)$ | M1
$= \left(\frac{1}{6}\right)^{-1} \times \frac{73}{216}$ | M1 | recognise product of roots
$= \frac{73}{36}$ | A1 | | [3]
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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 $\frac { ( \beta + 1 ) ( \gamma + 1 ) } { ( \alpha + 1 ) ^ { 2 } } + \frac { ( \gamma + 1 ) ( \alpha + 1 ) } { ( \beta + 1 ) ^ { 2 } } + \frac { ( \alpha + 1 ) ( \beta + 1 ) } { ( \gamma + 1 ) ^ { 2 } } = \frac { 73 } { 36 }$.
\hfill \mbox{\textit{CAIE FP1 2010 Q7 [9]}}