1 The roots of the cubic equation \(2 x ^ { 3 } + x ^ { 2 } - 7 = 0\) are \(\alpha , \beta\) and \(\gamma\). Using the substitution \(y = 1 + \frac { 1 } { x }\), or otherwise, find the cubic equation whose roots are \(1 + \frac { 1 } { \alpha } , 1 + \frac { 1 } { \beta }\) and \(1 + \frac { 1 } { \gamma }\), giving your answer in the form \(a y ^ { 3 } + b y ^ { 2 } + c y + d = 0\), where \(a , b , c\) and \(d\) are constants to be found.

## Question 1:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $y = 1 + \frac{1}{x} \Rightarrow x = \frac{1}{y-1}$ | M1 | |
| $\frac{2}{(y-1)^3} + \frac{1}{(y-1)^2} - 7 = 0 \Rightarrow 2 + (y-1) - 7(y-1)^3 = 0$ | A1 | |
| $\Rightarrow 7(y^3 - 3y^2 + 3y - 1) - y + 1 - 2 = 0 \Rightarrow 7y^3 - 21y^2 + 20y - 8 = 0$ | M1A1 [4] | ALT METHOD: $\sum\alpha$, $\sum\alpha\beta$, $\alpha\beta\gamma$ M1 A1, $\sum(1+1/\alpha)$ etc M1 A1√ |

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1 The roots of the cubic equation $2 x ^ { 3 } + x ^ { 2 } - 7 = 0$ are $\alpha , \beta$ and $\gamma$. Using the substitution $y = 1 + \frac { 1 } { x }$, or otherwise, find the cubic equation whose roots are $1 + \frac { 1 } { \alpha } , 1 + \frac { 1 } { \beta }$ and $1 + \frac { 1 } { \gamma }$, giving your answer in the form $a y ^ { 3 } + b y ^ { 2 } + c y + d = 0$, where $a , b , c$ and $d$ are constants to be found.

\hfill \mbox{\textit{CAIE FP1 2016 Q1 [4]}}