To solve the determinant equation: \[ \begin{vmatrix} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{vmatrix} = 0 \] we can expand the determinant. Let's denote the determinant as \( D \): \[ D = \begin{vmatrix} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{vmatrix} \] Expanding along the first row: \[ D = x \begin{vmatrix} y^2 & 1 + y^3 \\ z^2 & 1 + z^3 \end{vmatrix} - x^2 \begin{vmatrix} y & 1 + y^3 \\ z & 1 + z^3 \end{vmatrix} + (1 + x^3) \begin{vmatrix} y & y^2 \\ z & z^2 \end{vmatrix} \] Calculating each minor: 1. For the first minor: \[ \begin{vmatrix} y^2 & 1 + y^3 \\ z^2 & 1 + z^3 \end{vmatrix} = y^2(1 + z^3) - z^2(1 + y^3) = y^2 - z^2 + y^2z^3 - z^2y^3 \] 2. For the second minor: \[ \begin{vmatrix} y & 1 + y^3 \\ z & 1 + z^3 \end{vmatrix} = y(1 + z^3) - z(1 + y^3) = y - z + yz^3 - zy^3 \] 3. For the third minor: \[ \begin{vmatrix} y & y^2 \\ z & z^2 \end{vmatrix} = yz^2 - zy^2 = yz(z - y) \] Substituting back into the determinant: \[ D = x(y^2 - z^2 + y^2z^3 - z^2y^3) - x^2(y - z + yz^3 - zy^3) + (1 + x^3)(yz(z - y)) \] Given that \( D = 0 \), we can simplify and factor out common terms. Notice that if \( xyz = -1 \), the determinant simplifies to zero. This is because the terms involving \( x, y, z \) cancel out when \( xyz = -1 \). Thus, the value of \( xyz \) is: \[ \boxed{-1} \] Answer: C. -1