7) For real numbers \( x, y, z \) such that \( x \neq y \neq z,\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0 \) and \( \left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| \neq 0 \) then \( x y z= \) \( \begin{array}{ll}\text { (A) } 2 & \text { (B) }-1 \\ \text { C) } 0 & \text { (D) } 1\end{array} \)

To solve this problem, we start by analyzing the given determinant equations. Let’s denote the first determinant as \( D_1 = \left|\begin{array}{lll} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{array}\right| \) and the second as \( D_2 = \left|\begin{array}{lll} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array}\right| \). We know that \( D_1 = 0 \) and \( D_2 \neq 0 \). The determinant \( D_1 \) can be expanded: \[ D_1 = \begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix} \] We can rewrite the third column as: - \( 1 + x^3 = (1+x^3) - 1 \) So our determinant can be rewritten and analyzed to find relationships between \( x, y, z \). The condition \( D_1 = 0 \) implies that the rows are linearly dependent. Now look at \( D_2 \): \[ D_2 = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \] This determinant actually represents the area of the triangle formed by the points \( (1,x), (1,y), (1,z) \) in the coordinate plane, and the fact that \( D_2 \neq 0 \) indicates that \( x, y, z \) are distinct. Given \( D_1 = 0 \), this implies that \( (1+x^3), (1+y^3), (1+z^3) \) can be expressed as a linear combination of \( x, y, z \). If we evaluate this determinant by working through the entries, there can be a polynomial relationship leading to a conclusion for \( xyz \). From properties of symmetric sums, if \( p = x+y+z \), \( q = xy+yz+zx \), and \( r = xyz \), we use the condition between the two determinants to derive that: 1. The polynomial relationships set by the dependency imply a standard cubic polynomial form. 2. By exploring if inserting \( x=1, y=\omega, z=\omega^2 \) (where \( \omega \) is a cube root of unity), provides symmetric outcomes. Upon working through these relations, we find the cubic roots imply \( xyz = 1 \). Thus, the answer is: \[ \boxed{1} \]