Prove that: a^2&a^2-(b-c)^2&bc ^2&b^2-(c-a)^2&ca ^2&c^2-(a-b)^2&ab =(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2)

Let L.H.S= a^2&a^2-(b-c)^2&bc ^2&b^2-(c-a)^2&ca ^2&c^2-(a-b)^2&ab = a^2&-(b-c)^2&bc ^2&-(c-a)^2&ca ^2&-(a-b)^2&ab [Applying C_2 C_2 - C1] =(-1) a^2&(b-c)^2&bc ^2&(c-a)^2&ca ^2&(a-b)^2&ab =- a^2&b^2+c^2&bc ^2&c^2+a^2&ca ^2&a^2+b^2&ab [Applying C_2 C_2 - 2C_1] =- a^2+b^2+c^2&b^2+c^2&bc ^2+c^2+a^2&c^2+a^2&ca ^2+a^2+b&a^2+b^2&ab [Applying C_1 C_1 + C_2] =-(a^2+b^2+c^2) 1&b^2+c^2&bc 1&c^2+a^2&ca 1&a^2+b^2&ab =-(a^2+b^2+c^2) 1&b^2+c^2&bc 0&a^2-b^2&c(a-b) 0&a^2-c^2&b(a-c) =-(a^2+b^2+c^2)(a-b)(a-c) 1&b^2+c^2&bc 0&a+b&c 0&a+c&b [Taking (a - b) common from R_2 and (a - c) common from R_3] =-(a^2+b^2+c^2)(a-b)(c-a) 1 a+b&c +c&b [ (c-a)=-(a-c)] [Expanding along C_1] =-(a^2+b^2+c^2)(a-b)(c-a)(ab+b^2-ac-c^2) =-(a^2+b^2+c^2)(a-b)(c-a) a(b-c)+(b+c)(b-c) =(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2) =R.H.S Hence prove.

Prove that: a^2&a^2-(b-c)^2&bc ^2&b^2-(c-a)^2&ca ^2&c^2-(a-b)^2&a…

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Question

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

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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=\begin{vmatrix}\text{a}^2&-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - C1]$
$=(-1)\begin{vmatrix}\text{a}^2&(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=-\begin{vmatrix}\text{a}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - 2C_1]$
$=-\begin{vmatrix}\text{a}^2+\text{b}^2+\text{c}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2+\text{c}^2+\text{a}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2+\text{a}^2+\text{b}&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 + C_2]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\1&\text{c}^2+\text{a}^2&\text{ca}\\1&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}^2-\text{b}^2&\text{c}(\text{a}-\text{b})\\0&\text{a}^2-\text{c}^2&\text{b}(\text{a}-\text{c})\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{a}-\text{c})\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}+\text{b}&\text{c}\\0&\text{a}+\text{c}&\text{b}\end{vmatrix} $
$[$Taking $(a - b)$ common from $R_2$ and $(a - c)$ common from $R_3]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\times\left\{1\times\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{a}+\text{c}&\text{b}\end{vmatrix}\right\}$
$[\because(\text{c}-\text{a})=-(\text{a}-\text{c})] [$Expanding along $C_1]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})(\text{ab}+\text{b}^2-\text{ac}-\text{c}^2)$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\{\text{a}(\text{b}-\text{c})+(\text{b}+\text{c})(\text{b}-\text{c})\}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
Hence prove.

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