Question

If $\alpha ,\beta ,\gamma$ are roots of the equation $x^3+2x-3=0$, then the equation whose roots are $(\alpha -\beta )(\alpha -\gamma ),(\beta -\gamma )(\beta -\alpha ),(\gamma -\alpha )(\gamma -\beta )$ is:

• A. $6x^3+x^2+275=0$
• B. $x^3+6x^2+275=0$
• C. $x^3-6x^2-275=0$
• D. $x^3+6x^2-275=0$

Answer 1

The correct option is D $x^3+6x^2-275=0$

$x^3+2x-3=0$
$(x-1)(x^2+x+3)=0$
Hence,
Let $\gamma =1$
Now,
$\alpha +\beta =-1$ and $\alpha .\beta =3$
Sum of roots
$(\alpha -\beta )(\alpha -\gamma )+(\beta -\gamma )(\beta -\alpha )+(\gamma -\alpha )(\gamma -\beta )$
$=(\alpha -\beta )(\alpha -\gamma +\gamma -\beta )+(1-\alpha )(1-\beta )$
$=(\alpha -\beta {)}^2+1-(\alpha +\beta )+\alpha .\beta$
$=(\alpha +\beta {)}^2-4\alpha .\beta +1-(\alpha +\beta )+\alpha .\beta$
$=1-12+1-(-1)+3$
$=1-12+2+3$
$=-6$
Hence,
Sum of roots is $=\frac{-b}{a}=-6$
$\frac{b}{a}=6$
Since, $a=1$, $b=6$
Now product of roots is
$(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )(\beta -\alpha )(\gamma -\alpha )(\gamma -\beta )$
$=-(\alpha -\beta {)}^2(1-\alpha {)}^2(1-\beta {)}^2$
$=-(\alpha -\beta {)}^2[1-(\alpha +\beta )+\alpha .\beta {]}^2$
$=-\left[\right(\alpha +\beta {)}^2-4\alpha .\beta \left]\right[1-(\alpha +\beta )+\alpha .\beta {]}^2$
$=-[1-12][1+1+3{]}^2$
$=-(-11)\left(25\right)$
$=275$
$=\frac{-c}{a}$
Hence,
$c=-275$
Hence, the required equation is $x^3+6x^2-275=0$.