Question

The quadratic equation whose roots are $\left(\frac{\alpha}{\gamma}\right)^{3}$ and $\left(\frac{\beta}{\gamma}\right)^{3},$ where $\alpha, \beta, \gamma$ are roots of the equation $x^{3}-8=0,$ is

• A. $x^{2}+x+1=0$
• B. $x^{2}+2 x+4=0$
• C. $x^{2}-2 x+4=0$
• D. $x^{2}-2 x+1=0$

Answer 1

Answer: (D)

As $\alpha, \beta, \gamma$ are roots of $x^{3}-8=0$
$\therefore \quad x=2,2 \omega, 2 \omega^{2}$
$\therefore \alpha, \beta, \gamma$ are $2,2 \omega, 2 \omega^{2}$
$\therefore \alpha^{3}, \beta^{3}, \gamma^{3}$ is $8,8 \omega^{3}, 8 \omega^{6}$ or, $\alpha^{3}, \beta^{3}, \gamma^{3}$ is 8,8,8
Now, sum of the roots of the equation which is to be formed
$S=\frac{\alpha^{3}}{\gamma^{3}}+\frac{\beta^{3}}{\gamma^{3}}=1+1=2$
$P=$ product of roots $=\frac{\alpha^{3} \beta^{3}}{\gamma^{6}}=1 .$
$ \left(\because \alpha^{3}=\beta^{3}=\gamma^{3}\right)$
$\therefore \quad$ Required equation is $x^{2}-$ sum of roots $(x)+$ product of roots $=0$
$\Rightarrow x^{2}-2 x+1=0$