question (doubt)....
posted by MATHS .
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) ...