CAIE FP1 2006 November — Question 6

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2006
SessionNovember
TopicRoots of polynomials

6 The roots of the equation $$x ^ { 3 } + x + 1 = 0$$ are \(\alpha , \beta , \gamma\). Show that the equation whose roots are $$\frac { 4 \alpha + 1 } { \alpha + 1 } , \quad \frac { 4 \beta + 1 } { \beta + 1 } , \quad \frac { 4 \gamma + 1 } { \gamma + 1 }$$ is of the form $$y ^ { 3 } + p y + q = 0$$ where the numbers \(p\) and \(q\) are to be determined. Hence find the value of $$\left( \frac { 4 \alpha + 1 } { \alpha + 1 } \right) ^ { n } + \left( \frac { 4 \beta + 1 } { \beta + 1 } \right) ^ { n } + \left( \frac { 4 \gamma + 1 } { \gamma + 1 } \right) ^ { n }$$ for \(n = 2\) and for \(n = 3\).