solve to find x and y:-
x^3+8y^3+x+2y=0
try grouping
(x+2y)(x^2 - 2xu + 4y^2) + (x+2y) = 0
(x+2y)(x^2 - 2xy + 4y^2 + 1) = 0
x = -2y or x^2 -2xy + 4y^2 + 1 = 0
treat the second like a quadratic with
a=1
b=-2y
c=4y^2+1
x = (2y ± √(4y^2 - 4(1)(4y^2+1))/2
= (2y ± √(-12y^2 - 4)/2
notice the inside of the √ will always be negative so there is no real solution to that second part
so x = -2y
from that you can find an infinite number of solutions,
e.g. (-2,1), (4,-2), (-5.2 , 2.6) ...