Prove that:

Solution:

Let LHS $=\Delta=\mid \begin{array}{lll}a & a+b & a+2 b\end{array}$

$\begin{array}{lll}a+2 b & a & a+b \\ a+b & a+2 b & a \mid\end{array}$

$\Delta=\mid \begin{array}{lll}3 a+3 b & 3 a+3 b & 3 a+3 b\end{array}$

$\begin{array}{ccr}a+2 b & a & a+b \\ a+b & a+2 b & a \mid\end{array}$   $\left[\right.$ Applying $\left.\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}\right]$

$=3(a+b) \mid 1 \quad 1 \quad 1$

$\begin{array}{ccc}a+2 b & a & a+b \\ a+b & a+2 b & a\end{array}$  [Taking out $3(\mathrm{a}+\mathrm{b})$ common from $\mathrm{R}_{1}$ ]

$=3(a+b) \mid 0 \quad 0 \quad 1$

$\begin{array}{lll}2 b & -b & a+b\end{array}$

$-b \quad 2 b \quad a \mid \quad\left[\right.$ Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}$ and $\left.\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}\right]$

$=3(a+b) b^{2} \mid 0 \quad 0 \quad 1$

$\begin{array}{lll}2 & -1 & a+b\end{array}$

$\begin{array}{llll}-1 & 2 & a \mid & {\left[\text { Taking out } b \text { common from } \mathrm{C}_{1} \text { and } \mathrm{C}_{2} \text { ] }\right.}\end{array}$

$=3(a+b) b^{2} \times 3$

$=9(a+b) b^{2}$

$=$ RHS