For real numbers x, y and z,if x neq y neq z, | vedclass

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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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(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 third column: $\left|\begin{array}{ccc}x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1\end{array}ight| + \left|\begin{array}{ccc}x & x^2 & x^3 \ y & y^2 & y^3 \ z & z^2 & z^3\end{array}ight|=0$ In the second determinant,take $x, y, z$ common from rows $1, 2, 3$ respectively: $\left|\begin{array}{ccc}x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1\end{array}ight| + xyz \left|\begin{array}{ccc}1 & x & x^2 \ 1 & y & y^2 \ 1 & z & z^2\end{array}ight|=0$ For the first determinant,swap column $3$ with column $2$,then column $2$ with column $1$: $\left|\begin{array}{ccc}x & x^2 & 1 \ y & y^2 & 1 \ z & z^2 & 1\end{array}ight| = -\left|\begin{array}{ccc}x & 1 & x^2 \ y & 1 & y^2 \ z & 1 & z^2\end{array}ight| = \left|\begin{array}{ccc}1 & x & x^2 \ 1 & y & y^2 \ 1 & z & z^2\end{array}ight|$ Substituting this back: $(1 + xyz) \left|\begin{array}{ccc}1 & x & x^2 \ 1 & y & y^2 \ 1 & z & z^2\end{array}ight| = 0$ Since the second determinant is given as non-zero,we must have $1 + xyz = 0$,which implies $xyz = -1$.

For real numbers $x, y$ and $z$,if $x \neq y \neq z$,$\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $\left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right| \neq 0$,then $xyz = $ . . . . . . .

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