Solution #2

We want to find x³+y³. Factoring, we have:

x³+y³=(x+y)(x²-xy+y²)

Since x + y = 1, this is equal to x²-xy+y².

We then have x²-xy+y²=(x²+2xy+y²)-3xy

=(x+y)²-3xy

=1²-3*1

=-2

Alternatively, (x+y)³ = 1³=1. Expanding, (x+y)³ = x³ + 3x² + 3xy² + y³ = (x³+y³) + 3xy(x+y) = (x³+y³) + 3.
So (x³+y³) +3 = 1, and x³+y³=-2.