a + b + c = 6 → b + c = 6 - a (1)
ac + ab + bc = 5 → a ( b + c) = 5 - bc (2)
abc = -12 → bc = -12/a (3)
Put (1) and (3) into (2)
a ( 6 - a) = 5 - (-12/a)
6a - a^2 = 5 + 12/a multiply through by a
6a^2 - a^3 = 5a + 12 rearrange as
a^3 - 6a^2 + 5a + 12 = 0
Using the Rational Zeroes Theorem, 3 is shown to be a root
Using synthetic division, we can find the remaining polynomial
3 [ 1 - 6 5 12 ]
3 - 9 -12
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1 - 3 -4 0
The remaining polynomial is
a^2 - 3a - 4
The zeroes of this are
(a - 4) (a + 1) = 0
4 and -1
So...."a" can arbitrarily be either 3, 4 or -1
If we assign "a" as 4, "b" and "c" can be assigned as 4 and -1 respectively
And a^3 + b^3 + c^3 = 3^3 + 4^3 + (-1)^3 = 27 + 64 - 1 = 90
