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 $0$

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

$R_1\to R_1-R_2{R}_2\to R_3-R_3$

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

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

taking out common xy & yz from row1 and row 2 respectively