CAIE FP1 2019 November — Question 7

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
TopicRoots of polynomials

7 The equation \(x ^ { 3 } + 2 x ^ { 2 } + x + 7 = 0\) has roots \(\alpha , \beta , \gamma\).
  1. Use the relation \(x ^ { 2 } = - 7 y\) to show that the equation $$49 y ^ { 3 } + 14 y ^ { 2 } - 27 y + 7 = 0$$ has roots \(\frac { \alpha } { \beta \gamma } , \frac { \beta } { \gamma \alpha } , \frac { \gamma } { \alpha \beta }\).
  2. Show that \(\frac { \alpha ^ { 2 } } { \beta ^ { 2 } \gamma ^ { 2 } } + \frac { \beta ^ { 2 } } { \gamma ^ { 2 } \alpha ^ { 2 } } + \frac { \gamma ^ { 2 } } { \alpha ^ { 2 } \beta ^ { 2 } } = \frac { 58 } { 49 }\).
  3. Find the exact value of \(\frac { \alpha ^ { 3 } } { \beta ^ { 3 } \gamma ^ { 3 } } + \frac { \beta ^ { 3 } } { \gamma ^ { 3 } \alpha ^ { 3 } } + \frac { \gamma ^ { 3 } } { \alpha ^ { 3 } \beta ^ { 3 } }\).