Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

$\left|\begin{array}{ccc}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& 1+z^3\end{array}\right|\underset{R_1\to R_1-R_2}{\overset{R_2+R_2-R_3}{\to }}\left|\begin{array}{ccc}x-y& x^2-y^2& x^2+y^3\\ y-2& y^2-2^2& y^3-2^3\\ z& z^2& 1+z^3\end{array}\right|$

$\Rightarrow (x-y)(y-z)\left|\begin{array}{ccc}1& x+y& x^2+y^2+xy\\ 1& y+z& y^2+z^2+y^2\\ z& z^2& 1+z^3\end{array}\right|\stackrel{R_1\to R_1-R_2}{\to }$

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

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

$\to c_2\to c_2-z{c}_1$ & $c_3\to c_3-z{c}_2$

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

$\Rightarrow (x-y)(y-z)(x-z)(-1+2(y^2-xy-y^2\left)\right)$

$\Rightarrow (x-y)(y-z)(x-z)(-1-xyz)$

$\Rightarrow (x-y)(y-z)(z-x)(1+xyz)=0$

$\because x,y,z$ are all differnt so,$1+xyz=0$

$\Rightarrow$ This formula might be remembered for future use.