An algebra problem by K. J. W.

@K. J. W. Just to let you know, there are actually two possible solutions. We can have a = -1, b = 1/2 and c = 2, giving us a + b + c = 3/2 = 1.5, but we can also have a = b = c = -1, giving us a + b + c = -3, (which is what I plugged in on my first attempt).

In order for there to be a unique solution, you may want to specify that a, b and c must be unequal. Otherwise, good question. :)

EDIT: O.k., now for an actual solution.

Plugging in a, b and c in turn, we have the following three equations:

(i) $a^{3} + ab + c = b$ , (ii) $ab^{2} + b^{2} + c = c$ , and (iii) $ac^{2} + bc + c = a$ .

From (ii) we see that $(a + 1) * b^{2} = 0$ . Since we are given that the numbers must be non-zero we can conclude that $a = -1$ .

Now (i) becomes $-1 - b + c = b$ , and so $c = 2b + 1$ .

Next, (iii) becomes $-c^{2} + bc + c = -1$ . Now plug $c = 2b + 1$ into this last equation and simplify to find that $-2b^{2} - b + 1 = 0$ , which has solutions $b = -1$ and $b = 1/2$ .

Since a, b and c must be distinct we can rule out $b = -1$ . This leaves us with $b = 1/2$ , and $c = 2b + 1 = 2$ .

Finally, we have $a + b + c = -1 + (1/2) + 2 = \boxed{1.5}$ .

After substituting a,b, and c to the equation; it will arrive at three equations:

a^3 + ba + c = b <----- 1

ab^2 + b^2 + c = c <-----2

ac^2 + bc + c = a <-----3

Start solving on eqn no. 2;

ab^2 + b^2 + c = c

ab^2 + b^2 = 0

Factor:

b^2(a+1) = 0

b = 0 and a = -1, but b is absurd since the condition is nonzero.

Substitute a = -1 to eqn no. 1,

-1 - b + c = b

this becomes c = 2b + 1, then substitute this to equation no. 3;

-4b^2 - 4b - 1 + 2b^2 + b + 2b + 1 = -1

Combine like terms then rearrange:

-2b^2 - b + 1 = 0

Factor this trinomial,

(1-2b)(b+1) = 0

values for b are 1/2 and -1, but -1 is absurd since the condition is distinct.

Substitute b = 1/2 to c = 2b + 1, you'll get c = 2. Last, get their sum;

-1 + 1/2 + 2 = 3/2 or 1.5