Suppose x + y = a and x^3 + y^3 = b where a is not equal to 0. Find xy in terms of a and b.
x3 + y3 = (x + y)(x2 - xy + y2)
Since x3 + y3 = b ---> (x + y)(x2 - xy + y2) = b
Since x + y = a ---> a·(x2 - xy + y2) = b ---> x2 - xy + y2 = b/a (equation #1)
Since x + y = a ---> (x + y)2 = a2
and (x + y)2 = x2 + 2xy + y2 ---> x2 + 2xy + y2 = a2 (equation #2)
Combining equation #2 and equation #1: x2 + 2xy + y2 = a2
x2 - xy + y2 = b/a
Subtracting: 3xy = a2 - b/a
Rewriting: 3xy = (a3 - b)/a
Dividing: xy = (a3 - b) / (3a)