 # 1/(w+a)+1/(w+b)+1/(w+c)=2w2and1/(w2+a)+1/(w2+b)+1/(w2+c)=2wwhat is1/(a+1)+1/(b+1)+1/(c+1)=A)2B)-2C)w2D)wwhere w is a complex cube root of unity.  Badiuddin askIITians.ismu Expert
148 Points
12 years ago

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

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srichaitanya.

Regards,

12 years ago

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

3 years ago
Can you expalex me the step
a/a+w + b/b+w + c/c+w = 1
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3 years ago
W3 = 1 therefore w x w​= 1 therefore w​= 1/w --(1)
1/a+w + 1/b+w +  1/c+w = 2w​= 2/w. (From 1)

Now take conjugate of the above entire equation, we get (I am representing conjugate of w as “p" for convenience)
1/a+p + 1/b+p + 1/c+p = 2/p
Now w is -1/2 +root3/2 and w²  is -1/2 – root 3/2 which is just the same as conjugate of w
Therefore w² =p
So,
1/ a+w2 + 1/b+w2 + 1/c+w2 = 2/w2
So w and w2 are riits of equation 1/a+x + 1/b+x + 1/c+x = 2/x
Nowhow puthow x=1 to get the answer as 2.