Question

lf $\alpha ,\beta ,\gamma$ 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. $(x-7{)}^3-21(x-7{)}^2+972=0$
• B. $(x+1{)}^3-21(x+1{)}^2+972=0$
• C. $(x+7{)}^3-21(x+7{)}^2+972=0$
• D. $(x+7{)}^3-21(x+7{)}^2-400=0$

Answer 1

The correct option is A $(x-7{)}^3-21(x-7{)}^2+972=0$

As $\alpha ,\beta ,\gamma$ are roots of $x^3-7x+6=0$
We have
$s_1=\alpha +\beta +\gamma =0s_2=\alpha \beta +\alpha \gamma +\beta \gamma =-7s_3=\alpha \beta \gamma =-6$
Now
$\alpha \beta +\alpha \gamma +\beta \gamma =-7\Rightarrow \alpha \beta \gamma +\alpha {\gamma }^2+\beta {\gamma }^2=-7\gamma \Rightarrow {\gamma }^2(\alpha +\beta )=-7\gamma -\alpha \beta \gamma$
$\Rightarrow -{\gamma }^3=6-7\gamma \Rightarrow {\gamma }^3=7\gamma -6$
Let $x={(\alpha -\beta )}^2={(\alpha +\beta )}^2-4\alpha \beta ={\gamma }^2-4\alpha \beta$
$\Rightarrow x\gamma ={\gamma }^3-4\alpha \beta \gamma =7\gamma -6+24\Rightarrow \gamma =\frac{18}{x-7}$
Replacing $x\to \frac{18}{x-7}$ in given equation, we get
${\left(\frac{18}{x-7}\right)}^3-7\left(\frac{18}{x-7}\right)+6=0\Rightarrow {18}^3-7\cdot 18(x-7)+6{(x-7)}^3=0\Rightarrow {(x-7)}^3-21(x-7)+972=0$