Aspire Faculty ID #16206 · Topic: NIMCET 2011 · Just now
NIMCET 2011

Roots of $x^{2} - 2x + 4 = 0$ are $\alpha, \beta$. Compute $ \alpha^{6} + \beta^{6} $.

Solution

Roots: $ \alpha = 1 + i\sqrt{3},\ \beta = 1 - i\sqrt{3} = 2(\cos 60^\circ \pm i\sin 60^\circ) $ So: $ \alpha = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) $ $ \Rightarrow \alpha^{6} = 2^{6} (\cos 2\pi + i\sin 2\pi) = 64 $ Same for $\beta$. So $ \alpha^{6} + \beta^{6} = 64 + 64 = 128 $
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