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FM June 2021 p13 q02
4268
The cubic equation \(2x^3 - 4x^2 + 3 = 0\) has roots \(\alpha, \beta, \gamma\). Let \(S_n = \alpha^n + \beta^n + \gamma^n\).
(a) State the value of \(S_1\) and find the value of \(S_2\).
(b)
Express \(S_{n+3}\) in terms of \(S_{n+2}\) and \(S_n\).
Hence, or otherwise, find the value of \(S_4\).
(c) Use the substitution \(y = S_1 - x\), where \(S_1\) is the numerical value found in part (a), to find and simplify an equation whose roots are \(\alpha + \beta, \beta + \gamma, \gamma + \alpha\).
(d) Find the value of \(\frac{1}{\alpha + \beta} + \frac{1}{\beta + \gamma} + \frac{1}{\gamma + \alpha}\).
Solution
(a) The value of \(S_1\) is given by the sum of the roots: \(S_1 = \alpha + \beta + \gamma = 2\).
To find \(S_2\), use the relation \(S_2 = S_1^2 - 2(0) = 4\).
(b)(i) The recurrence relation is \(S_{n+3} = 2S_{n+2} - \frac{3}{2}S_n\).
(b)(ii) Using the recurrence relation, \(S_4 = 2S_3 - \frac{3}{2}S_1 = 2(2S_2 - \frac{3}{2}S_0) - \frac{3}{2}S_1 = 4\).