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. $x^2+2x+4=0$
• C. $x^2-2x+4=0$
• D. $x^2-2x+1=0$

Answer 1

Answer: (D)

As $\alpha ,\beta ,\gamma$ are roots of $x^3-8=0$
$\therefore x=2,2\omega ,2{\omega }^2$
$\therefore \alpha ,\beta ,\gamma$ 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 $=\frac{{\alpha }^3{\beta }^3}{{\gamma }^6}=1.$
$\left(\because {\alpha }^3={\beta }^3={\gamma }^3\right)$
$\therefore$ Required equation is $x^2-$ sum of roots $(x)+$ product of roots $=0$
$\Rightarrow x^2-2x+1=0$