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Sum of reciprocals (Posted on 2019-03-20) Difficulty: 3 of 5
Given

a+b+c+d=0
abcd=1
a3+b3+c3+d3=1983

Find 1/a+1/b+1/c+1/d

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution Comment 1 of 1
This was easier than I though.  I'll be using the Newton-Girard Formulas like I did in A PowerSum Puzzle.

Then from the given quantities S(1) = P(1) = 0, P(4) = 1, and S(3) = 1983.

Substitute into S(3) - S(2)*P(1) + S(1)*P(2) - 3*P(3) = 0 to get 1983 - S(2)*0 + 0*P(2) - 3*P(3), which simplifies to P(3) = -661.

1/a + 1/b + 1/c + 1/d can be expressed as P(3)/P(4), which by direct substitution equals -661.

  Posted by Brian Smith on 2019-03-20 12:31:27
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