If $p+q+r=0$, prove that the value of the determinant
$$ \Delta= \begin{vmatrix}
pa & qb &rc \\
qc & ra &pb\\
rb& pc & qa \\
\end{vmatrix} =-pqr \begin{vmatrix}
a & b &c \\
b & c &a\\
c& a & b \\
\end{vmatrix}$$
My Try:Since $p+q+r=0$ we have
$$a(p+q+r)+b(p+q+r)+c(p+q+r)=0$$ $\implies$
$$(ap+qc+rb)+(qb+ra+pc)+(rc+pb+qa)=0 \tag{1}$$
Now applying $C_1 \to C_1+C_2+C_3$ and then applying $R_1 \to R_1+R_2+R_3$ for $\Delta$ we get
$$\Delta= \begin{vmatrix}
0 & qb &rc \\
qc+ra+pb & ra &pb\\
rb+pc+qa& pc & qa \\
\end{vmatrix}$$
Any clue here?
Best Answer
By Sarrus' rule, you have$$\begin{vmatrix}pa&qb&rc\\qc&ra&pb\\rb&pc&qa\end{vmatrix}=prq(a^3+b^3+c^3)-abc(p^3+q^3+r^3).$$But\begin{align*}p^3+q^3+r^3&=\overbrace{(p+q+r)^3}^{\phantom{0}=0}-3(p q+r q+p r) (\overbrace{p+q+r}^{\phantom{0}=0})+3 p q r\\&=3pqr\end{align*} and therefore your determinant is equal to \begin{align*}pqr(a^3+b^3+c^3-3abc)&=-pqr(3abc-a^3-b^3-c^3)\\&=-pqr\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix},\end{align*}again by Sarrus' rule.