Question

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

Answer 1

Let $f\left(x\right)=x^2-2x+4=0$ has roots $\alpha ,\beta$, then the equation whose roots are ${\alpha }^6,{\beta }^6$ is
$f\left(x^{1/6}\right)=0\Rightarrow x^{1/3}-2x^{1/6}+4=0\Rightarrow x^{1/6}(x^{1/6}-2)=-4$
Cubing on both sides,
$x^{1/2}(x^{1/2}-8-6x^{1/6}(x^{1/6}-2\left)\right)=-64\Rightarrow x^{1/2}(x^{1/2}-8+24)=-64\Rightarrow x+16\sqrt{x}=-64\Rightarrow x+64=-16\sqrt{x}$
Squaring on both the sides,
$x^2+128x+{64}^2=256x\Rightarrow x^2-128x+{64}^2=0$
Therefore, sum of roots is,
${\alpha }^6+{\beta }^6=128$