Prove that:

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

$=\frac{1}{a b c}\left|\begin{array}{ccc}-a b c & a b^{2}+a b c & a c^{2}+a b c \\ a^{2} b+a b c & -a b c & c^{2} b+a b c \\ a^{2} c+a b c & b^{2} c+a b c & -a b c\end{array}\right| \quad\left[\right.$ Applying $R_{1} \rightarrow a R_{1}, R_{2} \rightarrow b R_{2}$ and $R_{3} \rightarrow c R_{3}$ and then dividing by $\left.a b c\right]$

$=\frac{a b c}{a b c}\left|\begin{array}{ccc}-b c & a b+a c & a c+a b \\ a b+b c & -a c & c b+a b \\ a c+b c & b c+a c & -a b\end{array}\right|$

[Taking out $a, b$ and $c$ common from the three columns]

$\left|\begin{array}{ccc}a b+b c+c a & a b+b c+c a & a b+b c+c a \\ a b+b c & -a c & c b+a b \\ a c+b c & b c+a c & -a b\end{array}\right|$ [Applying $\left.R_{1} \rightarrow R_{1}+R_{2}+R_{3}\right]$

$=(a b+b c+c a)\left|\begin{array}{ccc}1 & 1 & 1 \\ a b+b c & -a c & c b+a b \\ a c+b c & b c+a c & -a b\end{array}\right|$

$=\left(\begin{array}{c}a b+b c+c a \\ \end{array}\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & -(a b+b c+a c) & c b+a b \\ a c+b c+a b & b c+a c+a b & -a b\end{array}\right|$ [Applying $C_{1} \rightarrow C_{1}-C_{3}$ and $\left.C_{2} \rightarrow C_{2}-C_{3}\right]$

$=\left(\begin{array}{c}a b+b c+c a \\ 0\end{array}\right)\left|\begin{array}{ccc}0 & 1 \\ a c+b c+a b & b c+a c+a b & -a b\end{array}\right| \quad$ [Applying $C_{1} \rightarrow C_{1}-C_{3}$ and $\left.C_{2} \rightarrow C_{2}-C_{3}\right]$

$=(a b+b c+c a)\left|\begin{array}{cc}0 & -(a b+b c+a c) \\ a c+b c+a b & b c+a c+a b\end{array}\right|$

$=(a b+b c+c a)(a b+b c+a c)^{2}$

$=(a b+b c+c a)^{3}$

Hence proved.