# Prove that

Question:

Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c^{2} \\ a b & b^{2}+b c & c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

Solution:

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

Taking out common factors $a, b$, and $c$ from $\mathrm{C}_{1}, \mathrm{C}_{2}$, and $\mathrm{C}_{3}$, we have:

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

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we have:

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

Applying $R_{2} \rightarrow R_{2}+R_{1}$, we have:

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

Applying $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}+\mathrm{R}_{2}$, we have:

$\Delta=a b c\left|\begin{array}{ccc}a & c & a+c \\ a+b & b & a \\ 2 b & 2 b & 0\end{array}\right|$

$=2 a b^{2} c\left|\begin{array}{ccc}a & c & a+c \\ a+b & b & a \\ 1 & 1 & 0\end{array}\right|$

Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$, we have:

$\Delta=2 a b^{2} c\left|\begin{array}{ccc}a & c-a & a+c \\ a+b & -a & a \\ 1 & 0 & 0\end{array}\right|$

Expanding along $R_{3}$, we have:

\begin{aligned} \Delta &=2 a b^{2} c[a(c-a)+a(a+c)] \\ &=2 a b^{2} c\left[a c-a^{2}+a^{2}+a c\right] \\ &=2 a b^{2} c(2 a c) \\ &=4 a^{2} b^{2} c^{2} \end{aligned}

Hence, the given result is proved.