1/(w+a)+1/(w+b)+1/(w+c)=2w²

and

1/(w²+a)+1/(w²+b)+1/(w²+c)=2w

what is

1/(a+1)+1/(b+1)+1/(c+1)=

A)2

B)-2

C)w²

D)w

where w is a complex cube root of unity.

Dear srichaitanya

let f(z) =1/(z+a)+1/(z+b)+1/(z+c)  a function in z

    given f(w) =2w²

and        f(w²)=2w

clearly one of the possible function is

    f(z)=2z²

so f(1)=2

I wonder what is the basis for assuming that f(z) = z² here.

Multiplying the first relation by w, we get

w/w+a + w/w+b + w/w+c = 2

or a/w+a + b/w+b + c/w+c = 1

Similarly from the second relation, we get a/a+w² + b/b+w² + c/c+w² = 1

Thus, w and w² are the roots of a/(x+a) + b/(x+b) + c/(x+c) = 1

which simplifies to x³ - (ab+bc+ca)x -2abc = 0. This means that the sum of roots of this cubic is zero. Let the third root be r.

Then r + w+w² = 0 which means r = 1.

Thus a/a+1 + b/b+1 + c/c+1 = 1 or 1/a+1 + 1/b+1 + 1/c+1 = 2