If x x 2 1 + x 3 y y 2 1 + y 3 z z 2 1 + z 3 = 0 and x ,   y ,   z are all distinct, then x y z =

Given,

If $\left|\begin{array}{lll}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& 1+z^3\end{array}\right|=0$

$\left|\begin{array}{lll}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\left|\begin{array}{lll}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=0$

Taking $x,y,z$ from $C_1,C_2,C_3$ respectively 

$\left|\begin{array}{lll}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+xyz\left|\begin{array}{lll}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$

By interchanging the columns twice

$\left|\begin{array}{lll}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+xyz\left|\begin{array}{lll}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|=0$

$\left|\begin{array}{lll}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|\left(1+xyz\right)=0$

$R_1\to R_1-R_2 \& R_2\to R_2-R_3$

$\left|\begin{array}{lll}x-y& x^2-y^2& 1-1\\ y-z& y^2-z^2& 1-1\\ z& z^2& 1\end{array}\right|\left(1+xyz\right)=0$

$\left|\begin{array}{lll}x-y& \left(x-y\right)\left(x+y\right)& 0\\ y-z& \left(y-z\right)\left(y+z\right)& 0\\ z& z^2& 1\end{array}\right|\left(1+xyz\right)=0$

$\left(x-y\right)\left(y-z\right)\left|\begin{array}{lll}1& \left(x+y\right)& 0\\ 1& \left(y+z\right)& 0\\ z& z^2& 1\end{array}\right|\left(1+xyz\right)=0$

$R_1\to R_1-R_2$

$\left(x-y\right)\left(y-z\right)\left|\begin{array}{lll}1-1& \left(x+y-y-z\right)& 0\\ 1& \left(y+z\right)& 0\\ z& z^2& 1\end{array}\right|\left(1+xyz\right)=0$

$\left(x-y\right)\left(y-z\right)\left|\begin{array}{lll}0& \left(x-z\right)& 0\\ 1& \left(y+z\right)& 0\\ z& z^2& 1\end{array}\right|\left(1+xyz\right)=0$

$\left(x-y\right)\left(y-z\right)\left[\left(-x+z\right)\times 1\right]\left(1+xyz\right)=0$

$\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(1+xyz\right)=0$

$\Rightarrow 1+xyz=0 \left\{\because x\ne y\ne z\right\}$

Hence, $xyz=-1$