The sum of 3 real numbers is known to be zero. If the sum of their cubes is -1, what is their product equal to?
+129850 x + y + z = 0 x^3 + y^3 + z^3 = - 1
x + y = -z x^3 + y^3 = - 1 - z^3
(x + y)^2 = (-z)^2 (x + y) (x^2 - xy + y^2) = -1 - z^3
x^2 + y^2 + 2xy = z^2 (-z) ( z^2 - 2xy- xy) = -1 - z^3
x^2 + y^2 = z^2 - 2xy (-z) ( z^2 - 3xy) = -1 - z^3
-z^3 + 3xyz = - 1 - z^3
3xyz = - 1
xyz = -1/3 = answer
