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A special equation (Posted on 2020-02-19)

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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Partial(?) solution

| Comment 1 of 4

Based on the cyclical nature of the equations, I started by assuming that a = b = c. Then by substitution we get:

a^3 + 2a^2 - 1 = 0

This has three real roots:

a = -1

a = -phi

a = phi - 1

where phi is the "golden ratio", (1 + sqrt(5)) / 2.

Since I found solutions where a = b = c, I conclude that (a-b)(b-c)(c-a) = 0.

Posted by tomarken on 2020-02-19 13:38:31

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