# Prove that:

Question:

Prove that:

$\left|\begin{array}{lll}a^{2} & a^{2}-(b-c)^{2} & b c \\ b^{2} & b^{2}-(c-a)^{2} & c a \\ c^{2} & c^{2}-(a-b)^{2} & a b\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$

Solution:

Let LHS $=\Delta=\mid \mathrm{a}^{2} \quad \mathrm{a}^{2}-(\mathrm{b}-\mathrm{c})^{2} \quad \mathrm{bc}$

$b^{2} \quad b^{2}-(c-a)^{2} \quad c a$

$c^{2} \quad c^{2}-(a-b)^{2} \quad a b \mid$

$\Rightarrow \Delta=\mid \mathrm{a}^{2}-(\mathrm{b}-\mathrm{c})^{2} \quad \mathrm{bc}$

$\begin{array}{llll}\mathrm{b}^{2} & -(\mathrm{c}-\mathrm{a})^{2} & \mathrm{ca} & \\ \mathrm{c}^{2} & -(\mathrm{a}-\mathrm{b})^{2} & \mathrm{ab} \mid & {\left[\text { Applying } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\right]}\end{array}$

$=(-1) \mid \mathrm{a}^{2} \quad(\mathrm{~b}-\mathrm{c})^{2} \quad \mathrm{bc}$

$\begin{array}{lll}b^{2} & (c-a)^{2} & c a \\ c^{2} & (a-b)^{2} & a b \mid\end{array}$

$=-\mid a^{2} \quad b^{2}+c^{2} \quad b c$

$\begin{array}{lll}b^{2} & c^{2}+a^{2} & c a \\ c^{2} & a^{2}+b^{2} & a b \mid\end{array}$ [Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-2 \mathrm{C}_{1}$ ]

$=-\mid a^{2}+b^{2}+c^{2} \quad b^{2}+c^{2} \quad b c$

$\mathrm{b}^{2}+\mathrm{c}^{2}+\mathrm{a}^{2} \quad \mathrm{c}^{2}+\mathrm{a}^{2} \quad \mathrm{ca}$

$\mathrm{c}^{2}+\mathrm{a}^{2}+\mathrm{b}^{2} \quad \mathrm{a}^{2}+\mathrm{b}^{2} \quad \mathrm{ab} \mid \quad$ [Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}$ ]

$=-\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right) \mid 1 \quad \mathrm{~b}^{2}+\mathrm{c}^{2} \quad \mathrm{bc}$

$1 \quad \mathrm{c}^{2}+\mathrm{a}^{2} \quad \mathrm{ca}$

$1 \quad \mathrm{a}^{2}+\mathrm{b}^{2} \quad \mathrm{ab} \mid$

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

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

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

$\begin{array}{lll}0 & \mathrm{a}^{2}-\mathrm{b}^{2} & \mathrm{c}(\mathrm{a}-\mathrm{b}) \\ 0 & \mathrm{a}^{2}-\mathrm{c}^{2} & \mathrm{~b}(\mathrm{a}-\mathrm{c}) \mid\end{array}$

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

$\begin{array}{llcl}0 & \mathrm{a}+\mathrm{b} & \mathrm{c} & \\ 0 & \mathrm{a}+\mathrm{c} & \mathrm{b} \mid & {\left[\text { Taking }(\mathrm{a}-\mathrm{b}) \text { common from } \mathrm{R}_{2} \text { and }(\mathrm{a}-\mathrm{c}) \text { common from } \mathrm{R}_{3}\right]}\end{array}$

$=\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a) \times\{1 \times \mid a+b$ c

$\mathrm{a}+\mathrm{c} \quad \mathrm{b} \mid\} \quad[\because(\mathrm{c}-\mathrm{a})=-(\mathrm{a}-\mathrm{c})] \quad$ [Expanding along $C_{1}$ ]

$=\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left(a b+b^{2}-a c-c^{2}\right)$

$=\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\{a(b-c)+(b+c)(b-c)\}$

$=(a-b)(c-a)(b-c)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$

$=\mathrm{RHS}$

Hence proved.