# question (doubt)....

posted by .

solve to find x and y:-
x^3+8y^3+x+2y=0

• question (doubt).... -

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) ...

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