Discussion Forum

$\begin{bmatrix}1 & bc+ad&b^{2}c^{2}+a^{2}{d}^{2} \\1 & ca+bd & c^{2}a^{2}+b^{2}d^{2} \\ 1 & ab+cd & a^{2}b^{2}+c^{2}d^{2} \end{bmatrix}$=

Answer:

Explanation:

$\begin{vmatrix}1 & 1&1 \\bc+ad & ca+bd &ab+cd \\ b^{2}c^{2}+a^{2}{d}^{2} & c^{2}a^{2}+b^{2}d^{2} & a^{2}b^{2}+c^{2}d^{2} \end{vmatrix}$

 Applying $c_{1} \rightarrow  c_{1}-c_{2}$  and  $c_{2}= c_{2}-c_{3}$

$\begin{vmatrix}0 & 0 &1\\ c(b-a)+d(a-b) & c(a-d)+b(d-a)&ab+cd \\ c^{2}(b^{2}-a^{2})+d^{2}(a^{2}-b^{2}) & c^{2}(a^{2}-d^{2})+b^{2}(d^{2}-a^{2}) & a^{2}b^{2}+c^{2}d^{2} \end{vmatrix}$

$\Rightarrow 1[\left\{c(b-a)+d(a-b)\right\}\left\{c^{2}(a^{2}-d^{2})+b^{2}(d^{2}-a^{2})\right\}$

$\Rightarrow-[\left\{c(a-d)+b(d-a)\right\}\left\{c^{2}(b^{2}-a^{2})+d^{2}(a^{2}-b^{2})\right\}]$

 often after simplify we are getting

  (a-b)(a-c)(b-c)(b-d)(a-d)(c-d)