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Complex (Posted on 2018-12-22) Difficulty: 3 of 5
A = 1 + x3/3! + x6/6! + x9/9! + ...
B = x + x4/4! + x7/7! + x10/10! + ...
C = x2/2! + x5/5! + x8/8! + x11/11! + ...

Find the value of A3 + B3 + C3 - 3ABC.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (1 votes)

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Solution Surprising Comment 3 of 3 |
I suspect this is supposed to be done by using a^3+b^3+c^3-3abc = (a+b+c)(a+bw+cw^2) (a+bw^2+cw) where w is the complex cube root of unity (-1+isqrt(3))/2 - hence 'Complex'. But I used this method, more or less following earlier posts.

A = sum from 1 to infinity x^(3n-3)/(3n-3)! = (1/3 e^(-x/2) (e^((3x)/2) + 2 cos((sqrt(3)x)/2)))
B = sum from 1 to infinity x^(3n-2)/(3n-2)! = (1/3 e^(-x/2) (e^((3x)/2) - 2 sin(1/6(π - 3 sqrt(3)x))))
C = sum from 1 to infinity x^(3n-1)/(3n-1)! = (1/3 e^(-x/2) (e^((3x)/2) - 2 sin(1/6 (3 sqrt(3) x + π)))) and indeed when added these are the same as sum from 1 to infinity x^(n-1)/(n-1)! = e^x. 

Now we have A^3+B^3+C^3 = 1/9 (e^(3 x) + 2 e^(-(3 x)/2) cos((3 sqrt(3) x)/2) + 6) and -3ABC = 1/9 (-e^(3 x) - 2 e^(-(3 x)/2) cos((3 sqrt(3) x)/2) + 3).

When added these cancel nicely to produce the number 9/9, or 1, which is the required value.


Edited on January 17, 2019, 7:13 am
  Posted by broll on 2019-01-17 07:05:37

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