Prove that bc-a^2&ca-b^2&ab-c^2 -b^2&ab-c^2&bc-a^2 -c^2&bc-a^2&ca-b^2 is divisible by (a + b + c) and find the quotient.

= bc-a^2&ca-b^2&ab-c^2 -b^2&ab-c^2&bc-a^2 -c^2&bc-a^2&ca-b^2 [Applying C_1 _1-C_2 and C_2 _2-C_3 ] = bc-a^2-ca+b^2&ca-b^2-ab +c^2&ab-c^2 -b^2-ab+c^2&ab-c^2-bc+a^2&bc-a^2 -c^2-bc+a^2&bc-a^2-ca+b^2&ca-b^2 = (b-a)(a+b+c)&(c-b)(a+b+c)&ab-c^2c-b)(a+b+c)&(a-c)(a+b+c)&bc-a^2a-c)(a+b+c)&(b-a)(a+b+c)&ca-b^2 [Taking (a+b+c) common from C_1 and C_2 each ] =(a+b+c)= b-a&c-b &ab-c^2 -b&a-c&bc-a^2 -c&b-a&ca-b^2 [Applying R_1 _1+R_2+R_3 ] =(a+b+c)= 0&0&ab+bc+ca-(a^2+b^2+c^2) -b&a-c&bc-a^2 -c&b-a&ca-b^2 [Expanding along R_1] =(a+b+c)^2 [ab+bc+ca-(a^2+b^2+c^2) ] [(c-b)(b-a)(a-c)^2 ] =(a+b+c)^2(ab+bc+ca-a^2-b^2-c^2) (bc-ac-b^2+ab-a^2-c^2+2ac) =(a+b+c) [(a+b+c)(a^2+b^2+c^2-ab-bc-ca)^2 ] Hence, given deteminant is divisible by (a + b + c) and quotient is (a+b+c) (a^2+b^2+c^2-ab-bc-ca )^2

Prove that bc-a^2&ca-b^2&ab-c^2 -b^2&ab-c^2&bc-a^2 -c^2&bc-a^2&ca…

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Question

Prove that $\begin{vmatrix}\text{bc}-\text{a}^2&\text{ca}-\text{b}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2&\text{ab}-\text{c}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2&\text{bc}-\text{a}^2&\text{ca}-\text{b}^2\end{vmatrix}$ is divisible by $(a + b + c)$ and find the quotient.

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Answer

$\Delta=\begin{vmatrix}\text{bc}-\text{a}^2&\text{ca}-\text{b}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2&\text{ab}-\text{c}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2&\text{bc}-\text{a}^2&\text{ca}-\text{b}^2\end{vmatrix}$
$\big[$Applying $\text{C}_1\rightarrow\text{C}_1-\text{C}_2$ and $\text{C}_2\rightarrow\text{C}_2-\text{C}_3\big]$
$\Delta=\begin{vmatrix}\text{bc}-\text{a}^2-\text{ca}+\text{b}^2&\text{ca}-\text{b}^2-\text{ab} +\text{c}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2-\text{ab}+\text{c}^2&\text{ab}-\text{c}^2-\text{bc}+\text{a}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2-\text{bc}+\text{a}^2&\text{bc}-\text{a}^2-\text{ca}+\text{b}^2&\text{ca}-\text{b}^2\\\end{vmatrix}$
$=\begin{vmatrix}(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})&(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})&\text{ab}-\text{c}^2\text{c}-\text{b})(\text{a}+\text{b}+\text{c})&(\text{a}-\text{c})(\text{a}+\text{b}+\text{c})&\text{bc}-\text{a}^2\text{a}-\text{c})(\text{a}+\text{b}+\text{c})&(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})&\text{ca}-\text{b}^2\end{vmatrix}$
$\big[$Taking $(\text{a}+\text{b}+\text{c})$ common from $\text{C}_1$ and $\text{C}_2$ each$\big]$
$\Delta=(\text{a}+\text{b}+\text{c})=\begin{vmatrix}\text{b}-\text{a}&\text{c}-\text{b} &\text{ab}-\text{c}^2\\\text{c}-\text{b}&\text{a}-\text{c}&\text{bc}-\text{a}^2\\\text{a}-\text{c}&\text{b}-\text{a}&\text{ca}-\text{b}^2\end{vmatrix}$
$\big[$Applying $\text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3\big]$
$\Delta=(\text{a}+\text{b}+\text{c})=\begin{vmatrix}0&0&\text{ab}+\text{bc}+\text{ca}-(\text{a}^2+\text{b}^2+\text{c}^2)\\\text{c}-\text{b}&\text{a}-\text{c}&\text{bc}-\text{a}^2\\\text{a}-\text{c}&\text{b}-\text{a}&\text{ca}-\text{b}^2\end{vmatrix}$
$[$Expanding along $R_1]$
$\Delta=(\text{a}+\text{b}+\text{c})^2\big[\text{ab}+\text{bc}+\text{ca}-(\text{a}^2+\text{b}^2+\text{c}^2)\big]\big[(\text{c}-\text{b})(\text{b}-\text{a})(\text{a}-\text{c})^2\big]$
$=(\text{a}+\text{b}+\text{c})^2(\text{ab}+\text{bc}+\text{ca}-\text{a}^2-\text{b}^2-\text{c}^2)\times(\text{bc}-\text{ac}-\text{b}^2+\text{ab}-\text{a}^2-\text{c}^2+2\text{ac})$
$=(\text{a}+\text{b}+\text{c})\big[(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})^2\big]$
Hence, given deteminant is divisible by $(a + b + c)$ and quotient is
$(\text{a}+\text{b}+\text{c})\big(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\big)^2$

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