If $x, y, z$ are different and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0,$ then show that $1+x y z=0$.
If $x, y, z$ are different and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0,$ then show that $1+x y z=0$.
We have $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|$
$=\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{lll}x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3}\end{array}\right|$
$=(-1)^{2}\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|+x y z\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| \quad\left(\text { Using } \mathrm{C}_{3} \leftrightarrow \mathrm{C}_{2} \text { and then } \mathrm{C}_{1} \leftrightarrow \mathrm{C}_{2}\right)$
$=\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|(1+x y z)$
$=(1+x y z)\left|\begin{array}{ccc}
1 & x & x^{2} \\
0 & y-x & y^{2}-x^{2} \\
0 & z-x & z^{2}-x^{2}
\end{array}\right| $ $( { Using } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1})$
Taking out common factor $(y-x)$ from $\mathrm{R}_{2}$ and $(z-x)$ from $\mathrm{R}_{3},$ we get
$\Delta=(1+x y z)(y-x)(z-x)\left|\begin{array}{ccc}
1 & x & x^{2} \\
0 & 1 & y+x \\
0 & 1 & z+x
\end{array}\right|$
$=(1+x y z)(y-x)(z-x)(z-y)$ (on expanding along $\mathrm{C}_{1}$ )
since $\Delta=0$ and $x, y, z$ are all different, i.e., $x-y \neq 0, y-z \neq 0, z-x \neq 0,$ we get $1+x y z=0$
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