Sum of 3 No.=0,the sum of their cubes = 99,then their product is

x^3+y^3+z^3 = (x+y+z)(x^2+y^^2+z^2- xy - xz - yz) + 3xyz
Then
99 = 0*(x^2+y^^2+z^2- xy - xz - yz) + 3xyz
xyz = 33

when the sum of 3 numbers is 0, the sum of their cubes = 3 times their product.............
therefore, sum of cubes = 3 (product of the numbers)
which implies the product is sum of cubes/3 = 99/3 =33

a+ b+ c = 0
so ( a+b) = - c
or ( a+b)^3 = - c^3
a^3 +b^3 + 3ab(a +b) = - c^3
a^3 + b^3 + c^3 = - 3ab(a+b)
or - 3ab*(-c) = 99
or abc = 33 answer