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(B) Given the determinant equation: $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \ y & y^2 & 1+y^3 \ z & z^2 & 1+z^3\end{array}ight|=0$.We can split t

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$-1$

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(B) Given the determinant equation: $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \ y & y^2 & 1+y^3 \ z & z^2 & 1+z^3\end{array}ight|=0$.We can split the determinant into two: $\left|\begin{array}{ccc}x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1\end{array}ight|+\left|\begin{array}{ccc}x & x^2 & x^3 \ y & y^2 & y^3 \ z & z^2 & z^3\end{array}ight|=0$.Taking $xyz$ common from the second determinant: $\left|\begin{array}{ccc}x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1\end{array}ight|+xyz\left|\begin{array}{ccc}1 & x & x^2 \ 1 & y & y^2 \ 1 & z & z^2\end{array}ight|=0$.By performing column swaps ($C_1 \leftrightarrow C_2$ then $C_2 \leftrightarrow C_3$),the second determinant becomes $\left|\begin{array}{ccc}x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1\end{array}ight|$.Thus,$(1+xyz)\left|\begin{array}{ccc}x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1\end{array}ight|=0$.Since $x 
eq y 
eq z$,the determinant $\left|\begin{array}{ccc}x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1\end{array}ight| 
eq 0$.Therefore,$1+xyz=0$,which implies $xyz = -1$.

If $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $x \neq y \neq z$,then $1+x y z$ is equal to

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