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∣a+b10a2−b2a−b1a3+b3a2+b2+aba2−b2∣=\begin{vmatrix} a+b & 1 & 0 \\ a^2-b^2 & a-b & 1 \\ a^3+b^3 & a^2+b^2+ab & a^2-b^2 \end{vmatrix} =

Solution

✅ Correct Option: 2

Apply R2→R2−(a−b)R1R_2 \to R_2 - (a-b)R_1, which makes R2=(0,0,1)R_2 = (0, 0, 1). Expanding along R2R_2: determinant =1⋅(−1)2+3[(a+b)(a2+b2+ab)−(a3+b3)]= 1 \cdot (-1)^{2+3}[(a+b)(a^2+b^2+ab) - (a^3+b^3)]. Since (a+b)(a2+b2+ab)=a3+b3+2ab(a+b)(a+b)(a^2+b^2+ab) = a^3+b^3+2ab(a+b), value =−2ab(a+b)= -2ab(a+b).

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