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

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Aicharya · Accepted Answer

Given, $\alpha, \beta$ are the roots of $x^2-2 x+4=0$

Now, $\begin{aligned} \alpha-\beta & =\sqrt{(\alpha+\beta)^2-4 \alpha \beta} \\ & =\sqrt{4-4 \times 4}=\sqrt{-12} \end{aligned}$

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, $\begin{aligned} \alpha^6+\beta^6 & =\left(-2 \omega^2\right)^6+(-2 \omega)^6 \\ & =64\left(\omega^3\right)^4+64\left(\omega^3\right)^2 \\ & =128 \quad\left[\because \omega^3=1\right] \end{aligned}$

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

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