1708\hspace{-16}$Let $\mathbf{f(x,y,z)=x^2.(y-z)+y^2.(z-x)+z^2.(x-y)}$\\\\ Now If We Put $\mathbf{x-y=\Leftrightarrow x=y}\;,$ We Get \\\\ $\mathbf{f(x,y,z)=0\Leftrightarrow (x-y)}$ is a factor of $\mathbf{f(x,y,z)}$\\\\ Similarly for $\mathbf{(y-z)}$ and $\mathbf{(z-x)}\;,$ We Get $\mathbf{f(x,y,z)=0}$\\\\ So $\mathbf{f(x,y,z)=0}$ has a factor $\mathbf{(x-y).(y-z).(z-x)}$\\\\ Now $\mathbf{x^2.(y-z)+y^2.(z-x)+z^2.(x-y)}$ has over all power is $\mathbf{3}$ in each term.\\\\ and $\mathbf{(x-y).(y-z).(z-x)}$ also have a power $3$\\\\ So $\mathbf{f(x,y,z)=x^2.(y-z)+y^2.(z-x)+z^2.(x-y)=K.(x-y).(y-z).(z-x)}$\\\\ Now for Calculation of $\mathbf{K}\;,$ Put $\mathbf{x=-1\;,y=0\;,z=+1}\;,$Get$\\\\ \mathbf{K=-1}$\\\\ So $\mathbf{f(x,y,z)=x^2.(y-z)+y^2.(z-x)+z^2.(x-y)=-(x-y).(y-z).(z-x)}$
1its the same method....different form
the expression within the modulus can be written in the form of a determinant
\begin{vmatrix} x^2 & x & 1\\ y^2 & y & 1\\ z^2 & z & 1 \end{vmatrix}
if we factorize now..put x=y;y=z;z=y the determinant reduces to 0 so x-y;y-z;z-x are factors
it can be written as k(x-y)(y-z)(z-x) [k is some non-zero constant]
\begin{vmatrix} x^2 & x & 1\\ y^2 & y & 1\\ z^2 & z & 1 \end{vmatrix} = k(x-y)(y-z)(z-x) now put x=0 on both sides and expand the determinant..u`l get an identity
k is coming as equal to -1....thus it is factorized
36but i also didn't focus on this...
x2 (y - z) + y2 (z - x) + z2 (x - y)
= x2y - x2z + y2z - y2x + z2x - z2y
= (x2y - z2y) + (z2x - x2z) + (y2z - y2x)
= y (x2 - z2) + xz (z - x) + y2 (z - x)
= - y (x + z)(z - x) + xz (z - x) + y2 (z - x)
= (z - x)(-xy - yz + xz + y2)
= (z - x)[x (z - y)- y (z - y)]
= (z - x)(z - y)(x - y)
= - (x - y)(y - z)(z - x) ;P
