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

(A) We have $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|$
$= \left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right| + \left|\begin{array}{lll}x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3}\end{array}\right|$
$= (-1)^{2} \left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| + x y z \left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| \quad (\text{Using } C_{3} \leftrightarrow C_{2} \text{ and then } C_{1} \leftrightarrow C_{2})$
$= (1+x y z) \left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|$
$= (1+x y z) \left|\begin{array}{ccc}1 & x & x^{2} \\ 0 & y-x & y^{2}-x^{2} \\ 0 & z-x & z^{2}-x^{2}\end{array}\right| \quad (\text{Using } R_{2} \rightarrow R_{2}-R_{1} \text{ and } R_{3} \rightarrow R_{3}-R_{1})$
Taking out common factors $(y-x)$ from $R_{2}$ and $(z-x)$ from $R_{3}$,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 $C_{1}$)
Since $\Delta=0$ and $x, y, z$ are all distinct,i.e.,$x-y \neq 0, y-z \neq 0, z-x \neq 0$,we get $1+x y z=0$.