Retarded Algebra

I've been working on this problem in each moment of my spare time for nearly 3 days straight. If I don't figure this shit out, I'll go crazy. I think there's something simple that I'm just not getting.

a + b + c = a^2 + b^2 + c^2 = a^3 + b^3 + c^3 = 3/2.

What is the value of a times b times c?

There is a solution of a very similar problem at http://blog.plover.com/math/symmetric-function.html

Basically, you can
(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 9/4

Subtracting a^2 + b^2 + c^2 from both sides gives

2ab + 2bc + 2ac = 3/4.     (*)

(a+b+c)^3 = a^3 + b^3 + c^3
          + 3a^2b + 3a^2c + 3b2^a + 3b^2c + 3c^2a + 3c^2b
       = 27/8

Subtracting a^3 + b^3 + c^3 from both sides gives

3a^2b + 3a^2c + 3b2^a + 3b^2c + 3c^2a + 3c^2b = 15/8    (**)

Now take (*)
and multiply it by a + b + c = 3/2, getting

2a^2b + 2a^2c + 2b2^a + 2b^2c + 2c^2a + 2c^2b + 6abc = 9/8

Multiply both sides by 3/2:

3a^2b + 3a^2c + 3b2^a + 3b^2c + 3c^2a + 3c^2b + 9abc = 27/16

Subtract (**) from this, leaving

9abc = -3/16

so

abc = -1/48.

It's quite possible that I fucked up the arithmetic somewhere, but the method I showed should work fine, as long as you do it carefully on a big sheet of paper.

Hope this helps.