Show that

$\Delta+\Delta_{1}=\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|+\left|\begin{array}{ccc}1 & 1 & 1 \\ y z & z x & x y \\ x & y & z\end{array}\right|$

$=\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|+\left|\begin{array}{lll}1 & y z & x \\ 1 & z x & y \\ 1 & x y & z\end{array}\right|$         [Interchanging rows and coloumns in $\Delta_{1}$ ]

$=\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|-\left|\begin{array}{lll}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|$        [Applying $C_{2} \leftrightarrow C_{3}$ in $\Delta_{1}$ ]

$=\left|\begin{array}{ccc}1 & x & x^{2} \\ 0 & y-x & y^{2}-x^{2} \\ 0 & z-x & z^{2}-x^{2}\end{array}\right|-\left|\begin{array}{ccc}1 & x & y z \\ 0 & y-x & z x-y z \\ 0 & z-x & x y-y z\end{array}\right|$       [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ ]

$=(y-x)(z-x)\left|\begin{array}{ccc}1 & x & x^{2} \\ 0 & 1 & y+x \\ 0 & 1 & z+x\end{array}\right|-(y-x)(z-x)\left|\begin{array}{ccc}1 & x & y z \\ 0 & 1 & -z \\ 0 & 1 & -y\end{array}\right|$        [Taking $(y-x)$ common from $R_{2}$ and $(z-x)$ common from $R_{3}$ ]

$=(y-x)(z-x)(z+x-y-x)-(y-x)(z-x)(-y+z) \quad$ [Expanding along first column]

$=(y-x)(z-x)(z-y)(1-1)$

$=0$

$\therefore \Delta+\Delta_{1}=0$