IF

(x+y)³ = x³ +y³ +3xy(x+y)

(x+y)³ = 5+3xy(x+y)..............(1)

now

x²+y²=3

(x²+y² )(x+y) = 3(x+y)

x³ +y³ +xy(x+y) = 3(x+y)

5+xy(x+y)= 3(x+y)

xy(x+y)= 3(x+y) -5.....(2)

now put value of xy(x+y) in(1) from (2)

(x+y)³ = 5+3( 3(x+y) -5 )

(x+y)³ -9(x+y) +10=0

let x+y =t

after solving this qubic equation

we get (t-2)(t²+2t+5) = 0

since t²+2t+5= (t+1)²+4>0 ( always positive)

so t-2=0

t=2 ==> x+y=2