If $x \neq y \neq z$ and $\left|\begin{array}{lll}x & x^{3} & x^{4}-1 \\ y & y^{3} & y^{4}-1 \\ z & z^{3} & z^{4}-1\end{array}\right|=0$, prove that $x y z(x y+y z+z x)=(x+y+z)$

If $x \neq y \neq z$ and $\left|\begin{array}{lll}x & x^{3} & x^{4}-1 \\ y & y^{3} & y^{4}-1 \\ z & z^{3} & z^{4}-1\end{array}\right|=0$ , prove that $x y z(x y+y z+z x)=(x+y+z)$

If $x \neq y \neq z$ and $\left|\begin{array}{lll}x & x^{3} & x^{4}-1 \\ y & y^{3} & y^{4}-1 \\ z & z^{3} & z^{4}-1\end{array}\right|=0$ , prove that $x y z(x y+y z+z x)=(x+y+z)$

Solution:

By properties of determinants, we can split the given determinant into 2 parts
$\rightarrow 0=\left|\begin{array}{lll} \mathrm{x} & \mathrm{x}^{3} & \mathrm{x}^{4} \\ \mathrm{y} & \mathrm{y}^{3} & \mathrm{y}^{4} \\ \mathrm{z} & \mathrm{z}^{3} & \mathrm{z}^{4} \end{array}\right|-\left|\begin{array}{lll} \mathrm{x} & \mathrm{x}^{3} & 1 \\ \mathrm{y} & \mathrm{y}^{3} & 1 \\ \mathrm{z} & \mathrm{z}^{3} & 1 \end{array}\right|$
Taking $x, y, z$ common from $R_{1}, R_{2}, R_{3}$ respectively
$\rightarrow 0=x y z\left|\begin{array}{lll} 1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3} \end{array}\right|-\left|\begin{array}{lll} x & x^{3} & 1 \\ y & y^{3} & 1 \\ z & z^{3} & 1 \end{array}\right|$
Operating $R_{1} \rightarrow R_{1}-R_{3}, R_{2} \rightarrow R_{2}-R_{3}$
$\begin{array}{l} \rightarrow 0=x y z\left|\begin{array}{ccc} 0 & x^{2}-z^{2} & x^{3}-z^{3} \\ 0 & y^{2}-z^{2} & y^{3}-z^{3} \end{array}\right|-\left|\begin{array}{ccc} x-z^{3}-z^{3} & 0 \\ y-z & y^{3}-z^{3} & 0 \\ 1 & z^{2} & z^{3} & z & z^{3} & 1 \end{array}\right| \\ \rightarrow\left|\begin{array}{ccc} x-z & (x-z)\left(x^{2}+x z+z^{2}\right) & 0 \\ y-z & (y-z)\left(y^{2}+y x+z^{2}\right) & 0 \\ z & z^{3} & 1 \end{array}\right|=x y z\left|\begin{array}{ccc} 0 & (x-z)(x+z) & (x-z)\left(x^{2}+x z+z^{2}\right) \\ 0 & (y-z)(y+z) & (y-z)\left(y^{2}+y z+z^{2}\right) \\ 1 & z^{2} & z^{3} \end{array}\right| \end{array}$
Taking $(x-z)$ and $(y-z)$ common from $R_{1}, R_{2}$
$\rightarrow(x-z)(y-z)\left|\begin{array}{ccc} 1 & \left(x^{2}+x z+z^{2}\right) & 0 \\ 1 & \left(y^{2}+y z+z^{2}\right) & 0 \\ z & z^{3} & 1 \end{array}\right|=(x-z)(y-z)\left|\begin{array}{ccc} 0 & x+z & \left(x^{2}+x z+z^{2}\right) \\ 0 & y+z & \left(y^{2}+y z+z^{2}\right) \\ 1 & z^{2} & z^{3} \end{array}\right|$
Expanding with $R_{3}$
$\rightarrow y^{2}+y z+z^{2}-x^{2}-x z-z^{2}=x y z\left(x y^{2}+x y z+x z^{2}+z y^{2}+y z^{2}+z^{3}-x^{2} y-x y z-y z^{2}-x^{2} z-x z^{2}-z^{3}\right)$
$\rightarrow(y-x)(y+x)+z(y-x)=x y z\left(x y^{2}+z y^{2}-x^{2} y-x^{2} z\right)$
$\begin{array}{l} \rightarrow(y-x)(x+y+z)=x y z\left(x y(y-x)+z\left(y^{2}-x^{2}\right)\right) \\ \rightarrow(y-x)(x+y+z)=x y z(x y(y-x)+z(x+y)(y-x)) \\ \rightarrow(y-x)(x+y+z)=x y z(x y(y-x)+(x z+y z)(y-x)) \\ \rightarrow(y-x)(x+y+z)=x y z(y-x)(x y+x z+y z) \\ \rightarrow x+y+z=x y z(x y+x z+y z) \end{array}$
Hence Proved