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(A) Given: $\begin{vmatrix} x & x^2 & 1+x^3 \ y & y^2 & 1+y^3 \ z & z^2 & 1+z^3 \end{vmatrix} = 0$ Using the property of determinants,we can split t

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(A) Given: $\begin{vmatrix} x & x^2 & 1+x^3 \ y & y^2 & 1+y^3 \ z & z^2 & 1+z^3 \end{vmatrix} = 0$ Using the property of determinants,we can split the third column: $\begin{vmatrix} x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & x^3 \ y & y^2 & y^3 \ z & z^2 & z^3 \end{vmatrix} = 0$ Taking $x, y, z$ common from the rows of the second determinant: $\begin{vmatrix} x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1 \end{vmatrix} + xyz \begin{vmatrix} 1 & x & x^2 \ 1 & y & y^2 \ 1 & z & z^2 \end{vmatrix} = 0$ Rearranging the columns of the second determinant to match the first: $\begin{vmatrix} x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1 \end{vmatrix} + xyz \begin{vmatrix} x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1 \end{vmatrix} = 0$ $(1 + xyz) \begin{vmatrix} x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1 \end{vmatrix} = 0$ The determinant $\begin{vmatrix} x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1 \end{vmatrix}$ is a Vandermonde-like determinant which equals $(x-y)(y-z)(z-x)$. Since $x, y, z$ are distinct,$(x-y)(y-z)(z-x) 
eq 0$. Therefore,$1 + xyz = 0$,which implies $xyz = -1$.

If $\left|\begin{array}{lll}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $x, y, z$ are all distinct,then $x y z=$

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