Determine all possible triplet(s) (a, b, c) of positive real numbers, with a ≤ b ≤ c, that satisfy the following system of equations:

a³ – 3a = b, and:

b³ – 3b = c, and:

c³ – 3c = a

(1) a^3-3a=b
(2) b^3-3b=c
(3) c^3-3c=a
(4) a<=b<=c

from 3,4 we get
c^3-3c<=c
c^3<=4c
c^2<=4
since c>0 then
c<=2

from 1,4 we have
a^3-3a>=a
a^3>=4a
a^2>=4
since a>0 then
a>=2
and since the smallest a can be is 2, and the largest c can be
is 2 and a<=c then we have a=c=2 and that makes b=2

thus the only solution is (2,2,2)

Posted by Daniel on 2009-05-16 13:42:28