Question

lf $x,y,z$ are all different and if $\left|\begin{array}{ccc}x& x^3& x^4-1\\ y& y^3& y^4-1\\ z& z^3& z^4-1\end{array}\right|=0$, then $xyz(xy+yz+zx)$ is equal to

• A. $0$
• B. $1$
• C. $x+y+z$
• D. $x^2{y}^2{z}^2$

Answer 1

Answer: (C)

$x+y+z$

$\left|\begin{array}{ccc}x& x^3& x^4-1\\ y& y^3& y^4-1\\ z& z^3& z^4-1\end{array}\right|=0$
$xyz\left|\begin{array}{ccc}1& x^2& x^3\\ 1& y^2& y^3\\ 1& z^2& z^3\end{array}\right|-1\left|\begin{array}{ccc}x& x^3& 1\\ y& y^3& 1\\ z& z^3& 1\end{array}\right|=0$
$R_1\to R_1-R_3,R_2\to R_2-R_3$
$(x-y)(y-z)xyz\left|\begin{array}{ccc}0& x+z& x^2+xz+z^2\\ 0& y+z& y^2+yz+z^2\\ 1& z^2& z^3\end{array}\right|-(x-y)(y-z)\left|\begin{array}{ccc}1& x^2+xz+z^2& 0\\ 1& y^2+yz+z^2& 0\\ z& z^3& 1\end{array}\right|=0$
$\Rightarrow xyz\left|\begin{array}{ccc}0& x+z& x^2+xz+z^2\\ 0& y+z& y^2+yz+z^2\\ 1& z^2& z^3\end{array}\right|-\left|\begin{array}{ccc}1& x^2+xz+z^2& 0\\ 1& y^2+yz+z^2& 0\\ z& z^3& 1\end{array}\right|=0$ ( $\because x\ne y\ne z$)
$R_1\to R_1-R_2$
$xyz(x-y)\left|\begin{array}{ccc}0& 1& (x+y+z)\\ 0& y+z& y^2+yz+z^2\\ 1& z^2& z^3\end{array}\right|-(x-y)\left|\begin{array}{ccc}0& (x+y+z)& 0\\ 1& y^2+yz+z^2& 0\\ z& z^3& 1\end{array}\right|=0$
$\Rightarrow -xyz(x-y)(xy+yz+xz)+(x-y)(x+y+z)=0$
$\Rightarrow xyx(xy+yz+xz)=x+y+z$ ($\because x\ne y$)