Question
Prove that:
$\begin{vmatrix}(\text{b}+\text{c})^2&\text{a}^2&\text{bc}\\\text{c}+\text{a})^2&\text{b}^2&\text{ca} \\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$ $=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$

Answer

$\text{L.H.S}=\begin{vmatrix}(\text{b}+\text{c})^2&\text{a}^2&\text{bc}\\\text{c}+\text{a})^2&\text{b}^2&\text{ca}\\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=\begin{vmatrix}(\text{b}+\text{c})^2-(\text{c}+\text{a})^2&\text{a}^2-\text{b}^2&\text{bc}-\text{ca}\\\text{c}+\text{a})^2-(\text{a}+\text{b})^2&\text{b}^2-\text{c}^2&\text{ca}-\text{ab}\\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_1]$
$=\begin{vmatrix}(\text{b}+\text{c})(\text{b}+2\text{c}+\text{a})&(\text{a}+\text{b})(\text{a}-\text{b})&\text{c}(\text{b}-\text{a})\\\text{c}-\text{a})(\text{b}+2\text{a}+\text{c})&(\text{b}-\text{c})(\text{b}+\text{c})&\text{a}(\text{c}-\text{b})\\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-(\text{b}+2\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-(\text{b}+2\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\\\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$[$Applying $x_2 - y_2 = (x + y)(x - y)$ and taking out $(a - b)$ common from $R_1$ and $(b - c)$ from $R_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\\\text{a}+\text{b})^2-\text{c}^2&\text{c}^2&\text{ab}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 - C_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\\\text{a}+\text{b}+\text{c})(\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
$[$Applying $x^2 - y^2 = (x + y)(x - y)$ in $C_1]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\-2&\text{b}+\text{c}&-\text{a}\\\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
$[$Taking out $(a + b + c)$ common from $C_1]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&\text{c}-\text{a}&\text{c}-\text{a}\\\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
$[$ Applying $R_2 \rightarrow R_2 - R_1]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&1&1\\\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
$[$Taking out $(c - a)$ common from $R_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&-\text{c}\\0&0&1\\\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}&\text{ab}\end{vmatrix}$
$[$Applying $C_2 \rightarrow C_2 - C_3]$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a}) \left\{(-1)\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&\\\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}\end{vmatrix}\right\} [$Expanding along $R_2]$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\{-2\text{c}^2+2\text{ab}-\text{a}^2-\text{b}^2-2\text{ab}+\text{c}^2\}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})(-\text{a}^2-\text{b}^2-\text{c}^2)$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$

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