FM June 2025 p11 q02
4116
The cubic equation \(x^3 + 2x + 1 = 0\) has roots \(\alpha, \beta, \gamma\).
- Find a cubic equation whose roots are \(\alpha^3 - 1, \beta^3 - 1, \gamma^3 - 1\).
- Find the value of \((\alpha^3 - 1)^2 + (\beta^3 - 1)^2 + (\gamma^3 - 1)^2\).
- Find the value of \((\alpha^3 - 1)^3 + (\beta^3 - 1)^3 + (\gamma^3 - 1)^3\).
Solution
(a) Let \(y = x^3 - 1\). Then \(x^3 = y + 1\). Substitute into the original equation:
\((y+1)^3 + 2(y+1) + 1 = 0\)
\(\Rightarrow 2(y+1) = -y - 2\)
\(8(y+1) = -(y+2)^3\)
\(\Rightarrow 8y + 8 = -y^3 - 6y^2 - 12y - 8\)
\(y^3 + 6y^2 + 20y + 16 = 0\)
(b) \((\alpha^3 - 1)^2 + (\beta^3 - 1)^2 + (\gamma^3 - 1)^2 = 6^2 - 2(20) = -4\)
(c) \((\alpha^3 - 1)^3 + (\beta^3 - 1)^3 + (\gamma^3 - 1)^3 = -6(-6) - 20(-6) - 16(3) = 96\)
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