Nonzero real numbers a, b, and c are such that

a²+b+c=1/a
b²+c+a=1/b
c²+a+b=1/c

Find (a-b)(b-c)(c-a)

(In reply to re: Partial(?) solution by Ady TZIDON)

It seems to me though, that tomarken's reasoning only holds, if we know that only one real value for (a-b)(b-c)(c-a) is possible. This is somehow implied by the puzzle, but not yet proven. And until proven otherwise, there may be solutions where a=b=c does not hold either.

Posted by JLo on 2020-02-23 10:37:28