CAIE FP1 2019 June — Question 6

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

6 The equation $$x ^ { 3 } - x + 1 = 0$$ has roots \(\alpha , \beta , \gamma\).
  1. Use the relation \(x = y ^ { \frac { 1 } { 3 } }\) to show that the equation $$y ^ { 3 } + 3 y ^ { 2 } + 2 y + 1 = 0$$ has roots \(\alpha ^ { 3 } , \beta ^ { 3 } , \gamma ^ { 3 }\). Hence write down the value of \(\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 }\).
    Let \(S _ { n } = \alpha ^ { n } + \beta ^ { n } + \gamma ^ { n }\).
  2. Find the value of \(S _ { - 3 }\).
  3. Show that \(S _ { 6 } = 5\) and find the value of \(S _ { 9 }\).