We are given the determinant equation: \[ \left| \begin{array}{ccc} x & x^2 & 2 + x^3 \\ y & y^2 & 2 + y^3 \\ z & z^2 & 2 + z^3 \end{array} \right| = 0 \] Step 1: Expand the Determinant
Consider the determinant expansion along the third column: \[ D = \begin{vmatrix} x & x^2 & 2 + x^3 \\ y & y^2 & 2 + y^3 \\ z & z^2 & 2 + z^3 \end{vmatrix}. \] We can rewrite the third column as: \[ \begin{bmatrix} 2 + x^3 \\ 2 + y^3 \\ 2 + z^3 \end{bmatrix} = \begin{bmatrix} x^3 \\ y^3 \\ z^3 \end{bmatrix} + \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}. \] Thus, we express the determinant as: \[ D = \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} + \begin{vmatrix} x & x^2 & 2 \\ y & y^2 & 2 \\ z & z^2 & 2 \end{vmatrix}. \] Step 2: Evaluate the Determinants
The first determinant, \[ \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} \] is a Vandermonde determinant, which evaluates to: \[ (x - y)(y - z)(z - x) xyz. \] The second determinant, \[ \begin{vmatrix} x & x^2 & 2 \\ y & y^2 & 2 \\ z & z^2 & 2 \end{vmatrix}, \] has all elements in the last column equal to 2. Factoring out the common term: \[ 2 \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix}. \] This is another Vandermonde determinant, evaluated as: \[ 2 (x - y)(y - z)(z - x). \] Since the given determinant is 0, we have: \[ (x - y)(y - z)(z - x) xyz + 2 (x - y)(y - z)(z - x) = 0. \] Factoring out \( (x - y)(y - z)(z - x) \), which is nonzero (as \( x, y, z \) are distinct), we get: \[ xyz + 2 = 0. \] Step 3: Solve for \( xyz \) \[ xyz = -2. \] Thus, the required value is: \( -2. \)