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```        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.
```
8 years ago

147 Points
```										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

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.  All the best srichaitanya.  Regards, Askiitians Experts Badiuddin
```
8 years ago
mycroft holmes
266 Points
```										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
```
7 years ago
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