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∣ab−cc+ba+cbc−aa−ba+bc∣\begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c \end{vmatrix} is equal to:

Solution

✅ Correct Option: 1

Direct expansion gives the standard identity (a+b+c)(a2+b2+c2)(a+b+c)(a^2+b^2+c^2).

A quick verification with a=1,b=1,c=0a=1, b=1, c=0: the determinant of [11111−1020]\begin{bmatrix}1&1&1\\1&1&-1\\0&2&0\end{bmatrix} is 44, and (a+b+c)(a2+b2+c2)=2×2=4(a+b+c)(a^2+b^2+c^2) = 2\times 2 = 4, while the other options give 00.

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