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Triple cube sum (Posted on 2019-06-28)

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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Some progress

| Comment 2 of 3 |

Expand each side of the second equation and simplify to yield a/b + b/c + c/a = -1.

Add 3 to each side of the third equation and then the left side can be factored to yield (a/b + b/c + c/a)*(b/a + c/b + a/c) = 1. Then b/a + c/b + a/c = -1.

Posted by Brian Smith on 2019-06-30 21:00:24

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