Question

If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$, 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 $x^2-2x+4=0$
$\therefore \alpha +\beta =2\dots$ (i)
and $\alpha \beta =4\dots$(ii)
Now, $\alpha -\beta =\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }$
$=\sqrt{4-4\times 4}=\sqrt{-12}$
$\Rightarrow \alpha -\beta =2\sqrt{3}i\dots$(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 ${\alpha }^6+{\beta }^6={\left(-2{\omega }^2\right)}^6+(-2\omega {)}^6$
$=64{\left({\omega }^3\right)}^4+64{\left({\omega }^3\right)}^2$
$=128\left[\because {\omega }^3=1\right]$