we can use the sum of two cubes formula:

Note that x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (x + y)(x^2 + y^2 - xy)

now we can plug in x+y= 1 and x^2+y^2=4

so x^3 + y^3 = (1)(4 - xy)

=> x^3 + y^3 = 4 - xy

if we wanted to, we could solve for x and y directly, but the question didn't ask for that

EDIT: Oh sorry, we're not done yet. We need to find the value of x^3 + y^3, so we don't want xy in the expression. let's find a numerical value for xy.

we are told x + y = 1

=> (x + y)^2 = 1^2

=> x^2 + 2xy + y^2 = 1 .........but x^2 + y^2 = 4

=> 4 + 2xy = 1

=> 2xy = -3

=> xy = -3/2

so x^3 + y^3 = 4 - xy = 4 + 3/2 = 11/2 or 5 1/2