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A special equation (Posted on 2020-02-19) Difficulty: 3 of 5
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)

No Solution Yet Submitted by Danish Ahmed Khan    
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Hints/Tips re: Partial(?) solution | Comment 2 of 4 |
(In reply to Partial(?) solution by tomarken)

Consider a polynom p= x^3+x^2-1

p=0 has at most 3 solutions, distinct or equal, call them a, b, c  - so: 
0=(x-a)*(x-b)*(x-c)  therefore AT LEAST ONE IS TRUE:
(x-a)=0  OR
(x-b)=0   OR
(x-c)=0 
  Easy to show that   a=b implies a=c (by substitution )
and since  NO 4th ANSWER IS POSSIBLE

We conclude that (a-b)(b-c)(c-a) = 0.

No need to solve the equation!




Edited on February 20, 2020, 5:59 am
  Posted by Ady TZIDON on 2020-02-20 05:24:41

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