Using properties of determinants prove that: $\left|\begin{array}{lll} b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2} \end{array}\right|=0$

Solution:

$=\left|\begin{array}{lll}
b(b-a) & b-c & c(b-a) \\
a(b-a) & a-b & b(b-a) \\
c(b-a) & c-a & a(b-a)
\end{array}\right|$
Taking (b-a) common from $\mathrm{C}_{1}, \mathrm{C}_{3}$
$=(b-a)^{2}\left|\begin{array}{lll}
b & b-c & c \\
a & a-b & b \\
c & c-a & a
\end{array}\right|$
Operating $R_{2} \rightarrow R_{2}-R_{1}+R_{3}$
$\begin{array}{l}
=\left|\begin{array}{lll}
\mathrm{b} & \mathrm{b}-\mathrm{c}-\mathrm{b}+\mathrm{c} & \mathrm{c} \\
\mathrm{a} & \mathrm{a}-\mathrm{b}-\mathrm{a}+\mathrm{b} & \mathrm{b} \\
\mathrm{c} & \mathrm{c}-\mathrm{a}-\mathrm{c}+\mathrm{a} & \mathrm{a}
\end{array}\right| \\
=(\mathrm{b}-\mathrm{a})^{2}\left|\begin{array}{lll}
\mathrm{b} & 0 & \mathrm{c} \\
\mathrm{a} & 0 & \mathrm{~b} \\
\mathrm{c} & 0 & \mathrm{a}
\end{array}\right|
\end{array}$
[Properties of determinants say that if 1 row or column has only 0 as its elements, the value of the determinant is 0 ]
$=0$
Hence Proved