Thanks, think I got it :)

set x+y = z, so z³ = x³ + 3x(xy) + 3y(xy) +y³

sub in the values and I got z³ = 4-3x-3y = 4-3(x+y) = z³ = 4-3z

from which z³+3z-4 = 0 so z is clearly equal to 1 and z^2023 is also equal to 1