Prove that:

Solution:

Let LHS $=\Delta=\mid \begin{array}{lll}1 & b+c & b^{2}+c^{2}\end{array}$

$\begin{array}{lcc}1 & c+a & c^{2}+a^{2}\end{array}$

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

$\left[\right.$ Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\left.\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\right]$

$=\mid \begin{array}{lll}0 & b-a & b^{2}-a^{2}\end{array}$

$0 \quad c-b \quad c^{2}-b^{2}$

$1 \quad a+b \quad a^{2}+b^{2}$

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

$\begin{array}{lll}0 & b-c & b^{2}-c^{2} \\ 1 & a+b & a^{2}+b^{2}\end{array}$

$\left[\right.$ Taking out $(-1)$ common from $\mathrm{R}_{1}$ and $\left.\mathrm{R}_{2}\right]$

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

$\begin{array}{lll}0 & 1 & b+c \\ 1 & a+b & a^{2}+b^{2}\end{array}$

$=(a-b)(b-c)\{1 \times \mid 1 \quad a+b$

$1 \quad \mathrm{~b}+\mathrm{c} \mid\} \quad$ [Expanding along $\mathrm{C}_{1}$ ]

$=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})$

$=\mathrm{RHS}$

$\Rightarrow \Delta=\mid 0 \quad(b+c)-(c+a) \quad\left(b^{2}+c^{2}\right)-\left(c^{2}+a^{2}\right) 0 \quad(c+a)-(a+b) \quad\left(c^{2}+a^{2}\right)-\left(a^{2}+b^{2}\right) 1$