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

The correct option is D $x^2-2x+1=0$

$x^3-8=0$
$(x-2)(x^2+2x+4)=0$
Hence
$x=2$ and $x=2w,2w^2$
Now let
$\gamma =2$ and $\alpha =2w$ $\beta =2w^2$
Hence
$(\frac{\alpha }{\gamma }{)}^3=(\frac{2w}{2}{)}^3=1$
$(\frac{\beta }{\gamma }{)}^3=(\frac{2w^2}{2}{)}^3=1$
Hence the roots are real and equal and are equal to 1.
Thus the equation will be
$(x-1{)}^2=0$
$x^2-2x+1=0$