FM June 2022 p12 q04
4235
The cubic equation \(2x^3 + 5x^2 - 6 = 0\) has roots \(\alpha, \beta, \gamma\).
(a) Find a cubic equation whose roots are \(\frac{1}{\alpha^3}, \frac{1}{\beta^3}, \frac{1}{\gamma^3}\).
(b) Find the value of \(\frac{1}{\alpha^6} + \frac{1}{\beta^6} + \frac{1}{\gamma^6}\).
(c) Find also the value of \(\frac{1}{\alpha^9} + \frac{1}{\beta^9} + \frac{1}{\gamma^9}\).
Solution
(a) Let \(y = x^{-3}\), then \(x = y^{-\frac{1}{3}}\). Substitute into the original equation:
\(2y^{-1} + 5y^{-\frac{2}{3}} - 6 = 0\)
Multiply through by \(y\) to eliminate the fractions:
\(5y^{-\frac{2}{3}} = 6 - 2y^{-1}\)
\(5y^{-\frac{2}{3}} = 6 - 2y^{-1}\)
\(125y = (6y - 2)^3\)
Expanding and simplifying gives:
\(216y^3 - 216y^2 - 53y - 8 = 0\)
(b) Using the identity \(\alpha^{-3} + \beta^{-3} + \gamma^{-3} = 1\) and \(\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3} = -\frac{53}{216}\), we find:
\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = (\alpha^{-3} + \beta^{-3} + \gamma^{-3})^2 - 2(\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3})\)
\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = 1^2 - 2\left(-\frac{53}{216}\right)\)
\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = \frac{161}{108}\)
(c) Using the equation \(216S_9 = 216S_6 + 53S_3 + 24\) and substituting the values from parts (a) and (b):
\(S_9 = \frac{399}{216} = \frac{133}{72}\)
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