For any positive real number x, find the value of (xaxb)a+b×(xbxc)b+c×(xcxa)c+a
(xaxb)a+b×(xbxc)b+c×(xcxa)c+a=(xa−b)a+b.(xb−c)b+c.(xc−a)c+a=xa2−b2.xb2−c2.xc2−a2=xa2−b2+b2−c2+c2−a2=x∘=1