| Exam Board | CAIE |
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | November |
| Topic | Roots 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 \(n = 3\).