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Triple cube sum (Posted on 2019-06-28) Difficulty: 3 of 5
a+b+c=6
(a-b)/c+(b-c)/a+(c-a)/b=(1+a/b)(1+b/c)(1+c/a)
a2/bc+b2/ca+c2/ab+ab/c2+bc/a2+ca/b2=-2
a3+b3+c3=?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution re: Some progress - Solution Comment 3 of 3 |
(In reply to Some progress by Brian Smith)

Continuing from where I left off, Multiply each of my two previous equations by abc to get a^2*b+b^2*c+c^2*a = -abc and a^2*c+b^2*a+c^2*b = -abc.


Cube each side of a+b+c=6 and arrange the terms to get (a^3+b^3+c^3) + 3*(a^2*b+b^2*c+c^2*a) + 3*(a^2*c+b^2*a+c^2*b) + 6*abc = 216.

Now substitute the prior equations into the last equation to yield (a^3+b^3+c^3) + 3*(-abc) + 3*(-abd) + 6*abc = 216, which simplifies to a^3+b^3+c^3 = 216.

  Posted by Brian Smith on 2019-07-02 18:57:25
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