Question

Let $a,b,c$ be the roots of the equation $p{x}^3+q{x}^2+rx+s=0$. If $\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b^2+ab& -ab\end{array}\right|=27,a+b+c\ge 0$ and $a^2+b^2+c^2=3,$ then the value of $3p+q$ is

Answer 1

Let
$\Delta =\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b^2+ab& -ab\end{array}\right|$

Multiplying $R_1,R_2,R_3$ by $a,b,c$ respectively,
and then taking $a,b,c$ common from $C_1,C_2,$ and $C_3,$ we get
$\Delta =\left|\begin{array}{ccc}-bc& ab+ac& ac+ab\\ ab+bc& -ac& bc+ab\\ ac+bc& bc+ac& -ab\end{array}\right|$

Now, using $R_1\to R_1+R_2+R_3$

$\Delta =\left|\begin{array}{ccc}ab+ac+bc& ab+ac+bc& ab+ac+bc\\ ab+bc& -ac& bc+ab\\ ac+bc& bc+ac& -ab\end{array}\right|$


$\Delta =(ab+bc+ca)\left|\begin{array}{ccc}1& 1& 1\\ ab+bc& -ac& bc+ab\\ ac+bc& bc+ac& -ab\end{array}\right|$

Now, applying $C_2\to C_1-C_2,$ and $C_3\to C_1-C_3,$ we get
$\Delta =(ab+bc+ca)\left|\begin{array}{ccc}1& 0& 0\\ ab+bc& ab+bc+ca& 0\\ ac+bc& 0& ab+bc+ca\end{array}\right|$

$\Delta =(ab+bc+ca{)}^3$
$\Rightarrow 27=(ab+bc+ca{)}^3\Rightarrow ab+bc+ca=3$

and $a^2+b^2+c^2=3,$
$\therefore (a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\Rightarrow a+b+c=\pm 3$
$\Rightarrow a+b+c=3(\because a+b+c\ge 0)$

So,
$a+b+c=-\frac{q}{p}=3$
$\Rightarrow 3p+q=0$