Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3+x^2+x+1=0$, then $(\alpha -\beta {)}^2+(\beta -\gamma {)}^2+(\gamma -\alpha {)}^2$, is

• A. $3$
• B. $-3$
• C. $4$
• D. $-4$

Answer 1

The correct option is D $-4$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+x^2+x+1=0$, then
$\alpha +\beta +\gamma =-1$, $\alpha \beta +\beta \gamma +\gamma \alpha =1$ and $\alpha \beta \gamma =-1$
$\alpha +\beta +\gamma =-1\Rightarrow {(\alpha +\beta +\gamma )}^2=1\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\gamma \alpha )=1\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=1-2=-1$
Therefore, ${(\alpha -\beta )}^2+{(\beta -\gamma )}^2+{(\gamma -\alpha )}^2$

$=2({\alpha }^2+{\beta }^2+{\gamma }^2)-2(\alpha \beta +\beta \gamma +\gamma \alpha )=2(-1)-2\left(1\right)=-4$