Are these polynomials?

$x^2+y^2+z^2-xy-yz-zx\\ = \dfrac{1}{2} (x^2-2xy+y^2+y^2-2yz+z^2+x^2-2xz+z^2) \\= \dfrac{(x-y)^2+(y-z)^2+(z-x)^2}{2}$

Similarly the denominator is $\dfrac{(a-b)^2+(b-c)^2+(c-a)^2}{2}$

$x-y=b-a$ , $y-z=c-b$ , $z-x = a-c$

So actually the numerator and the denominator are the same.

$\dfrac{ \dfrac{(x-y)^2+(y-z)^2+(z-x)^2}{2}}{\dfrac{(a-b)^2+(b-c)^2+(c-a)^2}{2}}=\dfrac{(x-y)^2+(y-z)^2+(z-x)^2}{(x-y)^2+(y-z)^2+(z-x)^2} = \boxed{1}$

the answer is sweet and simple "1" substitute b+c in place of x;c+a in place of y and a+b in place of z hence, ( (b+c)^{2} + (c+a)^{2} + (a+b)^{2} - (c+a)(a+b) - (a+b)(b+c) - (b+c)(c+a) ) / a^{2} + b^{2} + c^{2} -bc -ca- ab after simple solving, the result will come as.... a^{2} + b^{2} + c^{2} -bc -ca- ab / a^{2} + b^{2} + c^{2} -bc -ca- ab=1

moreover, (a-b)^{2} was given in order to confuse us!!