Question

If $\alpha ,\beta$ are the roots of the equation $x^2-2x+4=0$, then the equation whose roots are ${\alpha }^3,{\beta }^3$ is

• A. $x^2+8x+64=0$
• B. $x^2-8x+64=0$
• C. $x^2+16x+64=0$
• D. $x^2-16x+64=0$

Answer 1

The correct option is C $x^2+16x+64=0$

Let $f\left(x\right)=x^2-2x+4$, then $f\left(x\right)=0$ has roots as $\alpha ,\beta$
The equation whose roots are ${\alpha }^3,{\beta }^3$ is
$f\left(x^{1/3}\right)=0\Rightarrow x^{2/3}-2x^{1/3}+4=0\Rightarrow x^{1/3}(x^{1/3}-2)=-4$
Cubing on both sides,
$x(x-8-6x^{1/3}(x^{1/3}-2\left)\right)=-64\Rightarrow x(x-8-6(-4\left)\right)=-64\Rightarrow x(x+16)+64=0$

Hence, the required equation is,
$x^2+16x+64=0$