Question

If $x=b+c,y=c+a,z=a+b$, then find the value of $\frac{x^2+y^2+z^2-yz-zx-xy}{a^2+b^2+c^2-bc-ca-ab}$

Answer 1

Given, $x-y=(b+c)-(c+a)=b-a$
Similarly $y-z=c-b$ and $z-x=a-c$
Now $x^2+y^2+z^2-yz-zx-xy$
$=\frac{1}{2}[{(x-y)}^2+{(y-z)}^2+{(z-x)}^2]$
$=\frac{1}{2}[{(b-a)}^2+{(c-b)}^2+{(a-c)}^2]$
$=\frac{1}{2}[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]$
$a^2+b^2+c^2-bc-ca-ab$
$\therefore$ the value of the given expression of equal to $1$.