The value of left| {begin{array}{*{20}{c}} {{ | vedclass

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(C) Let $\Delta = \left| {\begin{array}{*{20}{c}} {{(b + c)}^2} & {{a^2}} & {{a^2}} \ {{b^2}} & {{(a + c)}^2} & {{b^2}} \ {{c^2}} & {{c^2}} & {{(a +

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$2abc(a + b + c)^3$

Vedclass · Answer

(C) Let $\Delta = \left| {\begin{array}{*{20}{c}} {{(b + c)}^2} & {{a^2}} & {{a^2}} \ {{b^2}} & {{(a + c)}^2} & {{b^2}} \ {{c^2}} & {{c^2}} & {{(a + b)}^2} \end{array}} ight|$.Applying $C_1 ightarrow C_1 - C_3$ and $C_2 ightarrow C_2 - C_3$:$\Delta = \left| {\begin{array}{*{20}{c}} {{(b + c)}^2 - a^2} & 0 & {{a^2}} \ 0 & {{(a + c)}^2 - b^2} & {{b^2}} \ {{c^2 - (a + b)^2}} & {{c^2 - (a + b)^2}} & {{(a + b)}^2} \end{array}} ight|$$\Delta = \left| {\begin{array}{*{20}{c}} {(b+c-a)(b+c+a)} & 0 & {{a^2}} \ 0 & {(a+c-b)(a+c+b)} & {{b^2}} \ {(c-a-b)(c+a+b)} & {(c-a-b)(c+a+b)} & {{(a + b)}^2} \end{array}} ight|$Taking $(a+b+c)$ common from $C_1$ and $C_2$:$\Delta = (a+b+c)^2 \left| {\begin{array}{*{20}{c}} {b+c-a} & 0 & {{a^2}} \ 0 & {a+c-b} & {{b^2}} \ {-(c-a-b)} & {-(c-a-b)} & {{(a + b)}^2} \end{array}} ight|$Expanding the determinant,we get $\Delta = 2abc(a+b+c)^3$.

The value of $\left| {\begin{array}{*{20}{c}} {{(b + c)}^2} & {{a^2}} & {{a^2}} \\ {{b^2}} & {{(a + c)}^2} & {{b^2}} \\ {{c^2}} & {{c^2}} & {{(a + b)}^2} \end{array}} \right|$ is equal to

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