Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

$\left|\begin{array}{ccc}x+1& x+2& x+a\\ x+2& x+3& x+b\\ x+3& x+4& x+c\end{array}\right|$$

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

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

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

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

$(-1{)}^2\left|\begin{array}{ccc}1& x& x^2\\ 2& y& y^2\\ 3& z& z^2\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$

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

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

$\therefore$ Correct answer $B$