$\left|\begin{array}{lll}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0, x \neq y \neq z$
$\left|\begin{array}{lll}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right|+\left|\begin{array}{lll}x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3\end{array}\right|=0$
$\left|\begin{array}{lll}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right|+\left|\begin{array}{lll}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right| x y z=0$
$C_1 \leftrightarrow C_2$ and $C_2 \leftrightarrow C_3$
$\left|\begin{array}{lll}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right|+\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| x y z=0$
$(1+x y z)\left|\begin{array}{lll}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right|=0$
$\because R_2 \rightarrow R_2-R_1$
$R_3 \rightarrow R_3-R_1$
$(1+x y z)\left|\begin{array}{ccc}x & x^2 & 1 \\ y-x & y^2-x^2 & 0 \\ z-x & z^2-x^2 & 0\end{array}\right|=0$
Expand with $r$ to $C_3$
$(1+x y z)$
$\{(y-x)(z-x)(z+x)-(z-x)(y-x)$ $(y+x)\}=0$
$(1+x y z)(y-x)(z-x)(z+x-y-x)=0$
$(1+x y z)(x-y)(y-z)(z-x)=0$
$\because \quad x \neq y \neq z \Rightarrow x y z+1=0$