Discussion
Let \(\Delta = \begin{vmatrix} x & y & 1 \\ x+y & y+1 & x+1 \\ 1 & x & y \end{vmatrix}\). If \(x+y=-1\), then the value of \(\Delta\) is equal to
Answer: D
The value of \(\Delta\) is 0.
Expanding the determinant along the first row and collecting terms gives
\[ \Delta = -(x^{3} + y^{3} - 3xy + 1) \]
This is \(-(x^{3} + y^{3} + c^{3} - 3xyc)\) with \(c = 1\), and the identity \(p^{3} + q^{3} + r^{3} - 3pqr = (p + q + r)(\ldots)\) shows it vanishes whenever the three numbers add to zero. Here \(x + y + 1 = 0\) because \(x + y = -1\). Directly: \(x^{3} + y^{3} = (x + y)^{3} - 3xy(x + y) = -1 + 3xy\), so the bracket is \(-1 + 3xy - 3xy + 1 = 0\).
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