NIMCET 2019 — Mathematics PYQ

Calculations:

Given, x, y, z are distinct real numbers and
$$\left|\begin{array}{ccc}x& amp;x^2& amp;2+x^3\\ y& amp;y^2& amp;2+y^3\\ z& amp;z^2& amp;2+z^3\end{array}\right|=0$$
To find the value of xyz, solve the determinant
$$<br>\left|\begin{array}{ccc}x& amp;x^2& amp;2+x^3\\ y& amp;y^2& amp;2+y^3\\ z& amp;z^2& amp;2+z^3\end{array}\right|=0$$

$x[y^2(2+z^3)-z^2(2+y^3)]-x^2[y(2+z^3)-z(2+y^3)]+(2+x^3)[y{z}^2-z{y}^2]=0$

$=2x{y}^2+xy{z}^3-2x{z}^2+x{y}^3{z}^2-2x^{2}y-x^{2}y{z}^3+2x^{2}z+x^2{y}^{3}z+2y{z}^2-2z{y}^2+x^{3}y{z}^2-x^3{y}^{2}z=0$

$=2xyz-2x{z}^2-2x^{2}y+2x^{2}z+2y^{2}z-2z{y}^2+x^{3}y{z}^2-x^3{y}^{2}z+x{y}^2{z}^2+x{y}^{2}z-x^{2}yz+x^{2}y{z}^2=0$

$=2(xyz-x{z}^2-x^{2}y+x^{2}z+y^{2}z-z{y}^2)+xyz(x{y}^2-x{z}^2-x^{2}y+x^{2}z+y{z}^2-z{y}^2)=0$

$=(2+xyz)(x{y}^2-x{z}^2-x^{2}y+x^{2}z+y{z}^2-z{y}^2)=0$

$(2+xyz)=0$

$xyz=-2$