# Prove that:

Question:

Prove that:

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

Solution:

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

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

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

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

$\begin{array}{llll}(\mathrm{a}+\mathrm{b})^{2} & \mathrm{c}^{2} & \mathrm{ab} \mid & {\left[\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}\right.}\end{array}$ and $\left.\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\right]$

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

$=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c}) \mid-(\mathrm{b}+2 \mathrm{c}+\mathrm{a}) \quad \mathrm{a}+\mathrm{b} \quad-\mathrm{c}$

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

$-(b+2 a+c) \quad b+c \quad-a$

$(a+b)^{2} \quad c^{2} \quad a b \mid \quad\left[\right.$ A pplying $x^{2}-y^{2}=(x+y)(x-y)$ and taking out $(a-b)$ co mmon from $R_{1}$ and $(b-c)$ from $\left.R_{2}\right]$

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

$-2(\mathrm{~b}+\mathrm{a}+\mathrm{c}) \quad \mathrm{b}+\mathrm{c} \quad-\mathrm{a}$

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

$=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|-2(\mathrm{~b}+\mathrm{c}+\mathrm{a}) \quad \mathrm{a}+\mathrm{b} \quad-\mathrm{c}-2(\mathrm{~b}+\mathrm{a}+\mathrm{c}) \quad \mathrm{b}+\mathrm{c} \quad-\mathrm{a}(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{a}+\mathrm{b}-\mathrm{c}) \quad \mathrm{c}^{2} \quad \mathrm{ab}\right|$

$\left[\right.$ Applying $x^{2}-y^{2}=(x+y)(x-y)$ in $\left.C_{1}\right]$

$=(a-b)(b-c)(a+b+c) \mid-2 \quad a+b \quad-c$

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

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

$\begin{array}{llll}(\mathrm{a}+\mathrm{b}-\mathrm{c}) & \mathrm{c}^{2} & \mathrm{ab} & {\left[\text { Taking out }(\mathrm{a}+\mathrm{b}+\mathrm{c}) \text { common from } \mathrm{C}_{1}\right]}\end{array}$

[Taking out $(\mathrm{a}+\mathrm{b}+\mathrm{c})$ common from $\mathrm{C}_{1}$ ]

$=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{a}+\mathrm{b}+\mathrm{c}) \mid-2 \quad \mathrm{a}+\mathrm{b} \quad-\mathrm{c} 0 \quad \mathrm{c}-\mathrm{a} \quad \mathrm{c}-\mathrm{a}(\mathrm{a}+\mathrm{b}-\mathrm{c}) \quad \mathrm{c}^{2}$

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

$=\left(\begin{array}{llllll}\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{c}-\mathrm{a}) \mid-2 & \mathrm{a}+\mathrm{b} & -\mathrm{c} 0 & 1 & 1(\mathrm{a}+\mathrm{b}-\mathrm{c}) & \mathrm{c}^{2}\end{array} \quad \mathrm{ab} \mid\right.$

[Taking out $(\mathrm{c}-\mathrm{a})$ common from $\mathrm{R}_{2}$ ]

$=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{c}-\mathrm{a}) \mid-2 \quad \mathrm{a}+\mathrm{b}+\mathrm{c} \quad-\mathrm{c} 0 \quad 0 \quad \mathrm{ab}$

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

$=(a-b)(b-c)(a+b+c)(c-a)\left\{(-1)\left|-2 \quad a+b+c(a+b-c) \quad c^{2}-a b \quad\right|\right\} \quad$ [Expanding along $\left.R_{2}\right]$

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

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

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

$=\mathrm{RHS}$