Question

show that (the above thing =(x-y) (y-z)(z-x).pls ans only if u kn this question​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO PROVE

$\displaystyle\begin{vmatrix} 1 & x + y & {x}^{2} + {y}^{2} \\ 1 & y + z & {y}^{2} + {z}^{2} \\ 1 & z + x & {z}^{2} + {x}^{2} \end{vmatrix} = (x - y)(y - z)(z - x) \: \:$

PROOF

$\displaystyle\begin{vmatrix} 1 & x + y & {x}^{2} + {y}^{2} \\ 1 & y + z & {y}^{2} + {z}^{2} \\ 1 & z + x & {z}^{2} + {x}^{2} \end{vmatrix}$

$= \displaystyle\begin{vmatrix} 1 & x + y & {x}^{2} + {y}^{2} \\ 0 & z - x & {z}^{2} - {x}^{2} \\ 0 & z - y & {z}^{2} - {y}^{2} \end{vmatrix} \big(R'_2=R_2-R_1 \: , R'_3=R_3-R_1 \big)$

$= \displaystyle\begin{vmatrix} 1 & x + y & {x}^{2} + {y}^{2} \\ 0 & z - x & (z + x)(z - x) \\ 0 & z - y & (z + y)(z - y) \end{vmatrix}$

$=(z - x)(z - y) \displaystyle\begin{vmatrix} 1 & x + y & {x}^{2} + {y}^{2} \\ 0 & 1 & (z + x) \\ 0 & 1 & (z + y) \end{vmatrix} \: \big(taking \: common \: from \: 2nd \: and \: 3rd \: row \big)$

$=(z - x)(z - y) (z + y - z - x) \: \: \big( \: on \: expansion \big)$

$=(z - x)(z - y) ( y - x)$

$= (x - y)(y - z)(z - x) \: \:$