Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3-7x+6=0$ then the equation whose roots are $(\alpha -\beta {)}^2,(\beta -\gamma {)}^2,(\gamma -\alpha {)}^2$ is

• A. $x^3-42x^2+400x-411$
• B. $x^3-42x^2+441x+400$
• C. $x^3-42x^2+441x-400$
• D. $x^3-42x^2-400x-411$

Answer 1

Answer: (C)

$x^3-42x^2+441x-400$

Let $\alpha ,\beta ,\gamma$ are the roots of the equation.
$x^3-7x+6=0$ ...........$\left(1\right)$
$\alpha +\beta +\gamma =0,\alpha \beta \gamma =-6$
Let, $y=(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$
$=(-\gamma {)}^2-4\left(\frac{-6}{\gamma }\right)$
$={\gamma }^2+\frac{24}{\gamma }$
$=x^2+\frac{24}{x}$
$\Rightarrow xy=x^3+24$
$\Rightarrow xy=7x-6+24$ [From $\left(1\right)$]
$\Rightarrow x(y-7)=18$
$\Rightarrow x=\frac{18}{(y-7)}$

Substituting $x=\frac{18}{(y-7)}$ in $x^3-7x+6$
$\Rightarrow {\left(\frac{18}{y-7}\right)}^3-7\left(\frac{18}{y-7}\right)+6=0$

$\Rightarrow (18{)}^3-7(18\left)\right(y-7{)}^2+6(y-7{)}^3=0$

$\Rightarrow 5832-126(y^2-14y+49)+6(y^3-21y^2+147y-343)=0$

$\Rightarrow y^3-42y^2+441y-400=0$

The equation with roots $(\alpha -\beta {)}^2,(\beta -\gamma {)}^2,(\gamma -\alpha {)}^2$ is
$x^3-42x^2+441x-400=0$