Of course, the solution can be obtained by substituting $x$ with $2$ .

My working is as follows:

$(x+y)^2 = x^2 + 2xy + y^2$

$\quad \Rightarrow (x+y)^2 = 4 + 2xy \quad ... (1)$

$(x^2+y^2) (x+y) = x^3 + xy^2 + x^2y +y^3$

$\quad \Rightarrow 4(x+y) = 8 + xy(x+y) \quad ... (2)$

$(1) \times (x+y) - (2) \times 2: \quad (x+y)^3 - 8(x+y) = 4(x+y) - 16$

$\quad \Rightarrow (x+y)^3 - 12(x+y) + 16 = 0$

Let $u = x+y$ ,

$\quad \Rightarrow u^3 -12u + 16= 0\quad ...(3)$

$(3): \quad (u-2)^2(u+4) = 0$

$\quad \Rightarrow u = x+y = \boxed{2}$ or $x +y = -4$ .

$x^{3}+y^{3}$ = $(x+y$ ) $(x^{2}+y^{2}-xy)$

$x^{2}+y^{2}=(4$

$so$

$(x+y)(4-xy)$ = $8$

$by\ options\ only$ $(x+y)$ = $2$

by dividing the two equations

((x^3)/(x^2))+((y^3)/(y^2))=8/4

x+y=2

x = 2 y = 0 is clearly solution

x+y=2