Question

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

Answer 1

Given: $\Delta =\left|\begin{array}{ccc}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& 1+z^3\end{array}\right|=0$
Here, expanding element of $C_3$ into two determinants.
$\Rightarrow \left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\underset{}{\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|}=0$
Taking $x,y,z$ common from $R_1,R_2,R_3$ respectively.
$\Rightarrow \underset{}{\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|}+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$
Replacing $C_3\leftrightarrow C_2$
$\Rightarrow (-1)\underset{}{\left|\begin{array}{ccc}x& 1& x^2\\ y& 1& y^2\\ z& 1& z^2\end{array}\right|}+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$
Replacing $C_1\leftrightarrow C_2$
$\Rightarrow (-1)(-1)\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|(1+xyz)=0$
Using $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$\Rightarrow \left|\begin{array}{ccc}1& x& x^2\\ 1-1& y-x& y^2-x^2\\ 1-1& z-x& z^2-x^2\end{array}\right|(1+xyz)=0$
$\Rightarrow \left|\begin{array}{ccc}1& x& x^2\\ 0& (y-x)& (y-x)(y+x)\\ 0& (z-x)& (z-x)(z+x)\end{array}\right|(1+xyz)=0$
Taking common factor $(y-x)$ from $R_2$ and $(z-x)$ from $R_3$.
$\Rightarrow (1+xyz)(y-x)(z-x)\left|\begin{array}{ccc}1& x& x^2\\ 0& 1& y+x\\ 0& 1& z+x\end{array}\right|=0$
$\Rightarrow (1+xyz)(y-x)(z-x)\left(1\right(z+x)-1(y+x\left)\right)=0$
$\Rightarrow (1+xyz)(y-x)\left(\right(z-x\left)\right(z+x-y-x)=0$
$\Rightarrow (1+xyz)(y-x)(z-x)(z-y)=0$
Since it is given that $x,y,z$ all are different,
i.e., $y-x\ne 0,z-x\ne 0,z-y\ne 0$
So, only possibility is
$(1+xyz)=0$
Hence proved.