The value of the determinant left|begin{arra | vedclass

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(C) Let $ \Delta = \left|\begin{array}{ccc}a-b & b+c & a \ b-c & c+a & b \ c-a & a+b & c\end{array}ight| $.Applying the column operation $ C_{3} \

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$ a^{3}+b^{3}+c^{3}-3abc $

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(C) Let $ \Delta = \left|\begin{array}{ccc}a-b & b+c & a \ b-c & c+a & b \ c-a & a+b & c\end{array}ight| $.Applying the column operation $ C_{3} ightarrow C_{3} + C_{2} $:$ \Delta = \left|\begin{array}{ccc}a-b & b+c & a+b+c \ b-c & c+a & a+b+c \ c-a & a+b & a+b+c\end{array}ight| $Taking $ (a+b+c) $ common from $ C_{3} $:$ \Delta = (a+b+c) \left|\begin{array}{ccc}a-b & b+c & 1 \ b-c & c+a & 1 \ c-a & a+b & 1\end{array}ight| $Applying $ R_{1} ightarrow R_{1} - R_{2} $ and $ R_{2} ightarrow R_{2} - R_{3} $:$ \Delta = (a+b+c) \left|\begin{array}{ccc}a-2b+c & b-a & 0 \ b-2c+a & c-b & 0 \ c-a & a+b & 1\end{array}ight| $Expanding along $ C_{3} $:$ \Delta = (a+b+c) [ (a-2b+c)(c-b) - (b-a)(b-2c+a) ] $$ \Delta = (a+b+c) [ (ac - ab - 2bc + 2b^{2} + c^{2} - bc) - (b^{2} - 2bc + ab - ab + 2ac - a^{2}) ] $$ \Delta = (a+b+c) [ ac - ab - 3bc + 2b^{2} + c^{2} - b^{2} + 2bc - 2ac + a^{2} ] $$ \Delta = (a+b+c) [ a^{2} + b^{2} + c^{2} - ab - bc - ac ] $$ \Delta = a^{3} + b^{3} + c^{3} - 3abc $.

The value of the determinant $ \left|\begin{array}{ccc}a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c\end{array}\right| $ is

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