FM June 2024 p13 q02
4172
The cubic equation \(x^3 + 2x^2 + 3x + 1 = 0\) has roots \(\alpha, \beta, \gamma\).
(a) Find a cubic equation whose roots are \(\alpha^2 + 1, \beta^2 + 1, \gamma^2 + 1\).
(b) Find the value of \((\alpha^2 + 1)^2 + (\beta^2 + 1)^2 + (\gamma^2 + 1)^2\).
(c) Find the value of \((\alpha^2 + 1)^3 + (\beta^2 + 1)^3 + (\gamma^2 + 1)^3\).
Solution
(a) Let \(y = \alpha^2 + 1\). Then \(\alpha^2 = y - 1\). Substitute into the original equation:
\((y - 1)^3 + 2(y - 1)^2 + 3(y - 1) + 1 = 0\)
Expanding and simplifying gives:
\(y^3 - y^2 + 4y - 5 = 0\)
(b) Using the result from part (a), we have:
\((\alpha^2 + 1)^2 + (\beta^2 + 1)^2 + (\gamma^2 + 1)^2 = 1 - 2(4) = -7\)
(c) Again using the result from part (a):
\((\alpha^2 + 1)^3 + (\beta^2 + 1)^3 + (\gamma^2 + 1)^3 = -7 - 4 + 5(3) = 4\)
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