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?
a+b+c=0
a^3+b^3+c^3=-1
a+b=-c
a^3+b^3 - (a+b)^3 = -1
(a+b)^3 = a^3 +b^3 +3abb +3aab
Cancel cubes
3aab + 3abb = -1
3ab(a+b)=-1
3ab(-c) =-1
3abc = 1
abc = 1/3
