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Using property of determinants show that \( \left|\begin{array}{ccc}(b+c)^{2} & a^{2} & a^{2} \\ b^{2} & (c+a)^{2} & b^{2} \\ c^{2} & c^{2} & (a+b)^{2}\end{array}\right|=2 a b c(a+b+c)^{3} \). |
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Answer» \(\begin{vmatrix}(b+c)^2&a^2&a^2\\b^2&(c+a)^2&c^2\\c^2&c^2&(a+b)^2\end{vmatrix}\) Applying C2→C2 - C1, C3→ C3 - C1 \(\begin{vmatrix}(b+c)^2&(a-(b+c))(a+b+c)&(a-(b+c))(a+b+c)\\b^2&(c+a-b)(c+a+b)&0\\c^2&0&(a+b-c)(a+b+c)\end{vmatrix}\) Applying C2 → \(\frac{C_2}{a+b+c}\), C3 → \(\frac{C_3}{a+b+c}\) = (a + b + c)2\(\begin{vmatrix}(b+c)^2&a-b-c&a-b-c\\b^2&c+a-b&0\\c^2&0&a+b-c\end{vmatrix}\) Applying R1 → R1 - R2 - R2 - R3 = (a + b + c)2\(\begin{vmatrix}2bc&-2c&-2b\\b^2&c+a-b&0\\c^2&0&a+b-c\end{vmatrix}\) Applying R1 → R2/2 = 2(a + b + c)2\(\begin{vmatrix}bc&-c&-b\\b^2&c+a-b&0\\c^2&0&a+b-c\end{vmatrix}\) Expanding determinant along row R3 = 2(a + b + c)2(c2(ab + bc - b2) + (a + b - c)(abc + bc2 - b2c + b2c)) = 2(a + b + c)2 c(abc + a2b + ab2) = 2abc (a + b + c)2(a + b + c) = 2abc (a + b + c)3 |
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