Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3+a{x}^2+bx+c=0$, then $\Sigma {\alpha }^2(\beta +\gamma )=$

• A. $ab+3c$
• B. $ab-3c$
• C. $3c-ab$
• D. $3abc$

Answer 1

Answer: (C)

$3c-ab$

As $\alpha ,\beta ,\gamma$ are the roots of $x^3+a{x}^2+bx+c=0$ gives
$\alpha +\beta +\gamma =-a\alpha \beta +\beta \gamma +\gamma \alpha =b\alpha \beta \gamma =-c$

Now $\sum {\alpha }^2(\beta +\gamma )=(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )-3\alpha \beta \gamma$

$=(-a)\left(b\right)-3(-c)=3c-ab$