[(b–c)/a + (c–a)/b + (a–b)/c].[a/(b–c) + b/(c–a) + c/(a–b)]

S1=(b-c)/a +(c-a)/b +(a-b)/c = [bc(b-c)+ca(c-a)+ab(a-b)]/abc

but  E=bc(b-c)+ca(c-a)+ab(a-b) can be factorised. if we put a=b

the expression E=0. so a-b is a factor. similarly b-c ,c-a are factors. so (a-b)(b-c)(c-a) is factor of E. E is of 3rd degree. so

E=K(a-b)(b-c)(c-a) putting a=0,b=1,c=2 we get K=-1. thus

E=-(a-b)(b-c)(c-a).

so S1=-(a-b)(b-c)(c-a)/abc.

S2= (a/(b-c) +b/(c-a) +c/(a-b))= [sigma[a(c-a)(a-b)]/(a-b)(b-c)(c-a)

but sigma[a(c-a)(a-b)]=sigma[a(a(b+c)-a^2-bc)]=sigma[-2a^3-abc]= -2(a^3+b^3+c^3)-3abc

but a^3+b^3+c^3=3abc,so sigma[a(c-a)(a-b)]=-9abc

so, S2= -9abc/(a-b)(b-c)(c-a)

S1*S2= 9