Question

If $\alpha$ and $\beta$ are the roots of $x^{2}-2 x+4=0$, then the value of $\alpha^{6}+\beta^{6}$ is

• A. 32
• B. 64
• C. 128
• D. 256

Answer 1

Answer: (C)

Given, $\alpha, \beta$ are the roots of $x^{2}-2 x+4=0$
$\therefore \alpha+\beta=2 \ldots$ (i)
and $\alpha \beta=4 \ldots$(ii)
Now, $\alpha-\beta=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}$
$=\sqrt{4-4 \times 4}=\sqrt{-12}$
$\Rightarrow \alpha-\beta=2 \sqrt{3}\, i \ldots$(iii)
On solving Eqs. (i) and (ii), we get
$\alpha=\frac{2+2 \sqrt{3}\, i}{2}=-2\left(\frac{-1+\sqrt{3}\, i}{2}\right)=-2 \omega^2$
and $\beta=\frac{2-2 \sqrt{3} i}{2}=-2\left(\frac{-1+\sqrt{3}\,i}{2}\right)=-2 \omega$
Now $\alpha^{6}+\beta^{6}=\left(-2 \omega^{2}\right)^{6}+(-2 \omega)^{6}$
$=64\left(\omega^{3}\right)^{4}+64\left(\omega^{3}\right)^{2}$
$=128\left[\because \omega^{3}=1\right]$