Intersection of Cubic curves This is the question which i am attempting to solve and it seems to be difficult to get rid of the exponents.

Show that a the two cubic curves $Y^3 = X^2 + X^3$ and $X^3 = Y^2 + Y^3$ intersect each other at nine points.

Any help would be appreciated. By the way i would also like to know whether there is any general method for solving such problems. 
 A: In addition to the usual generalities on Bezout's theorem and resultants, notice that for this system, $X^2 + Y^2 = 0$, so that there are 6 finite solutions (3 on each line $X = \pm iY$, organized as 1 transverse intersection point per line, plus one double point at (0,0) on each line, leading to a quadruple intersection of the curves at 0)  and accounting for the other 3 solutions, the solutions "at infinity" with $X^3=Y^3$, i.e., the asymptotic directions $(X/Y) \to \omega, \omega^3=1, |X| \to \infty$.  In other words, this example is built to show the need for complex projective geometry, and the counting of points according to their multiplicities, as the environment where the Bezout 3x3=9 calculation works.  
(the generalities are: if f(x,y)=0 and g(x,y)=0 are the polynomials, you can write down a basis for k[x,y]/(f=g=0), and check that as a vector space it is of dimension 9. If you meant that there are 9 distinct intersection points, you can check that by writing down the action of the multiply-by-x operator on this vector space and seeing whether it has repeated roots. This can all be done by hand but is not as efficient as what is done in computer algebra systems. )
A: Suppose $(x,y)$ satisfies both equations. Substituting one equation into the other, we obtain $x^2+y^2=0$. The only real solution to this equation are $x=y=0$, as pointed out by Moron.
Now suppose $x\neq 0$, and so $x$ is a complex number. Then $y = \pm i x$. Eliminating $y$ from the other equation yields two possibilities:


*

*$y = ix$, and $-ix^3 = x^2 + x^3$. We assume $x\neq 0$, so we have: $-ix = 1 + x$. This yields the solution: $x = (-1+i)/2, y=(-1-i)/2$.

*$y = -ix$, and $ix^3 = x^2 + x^3$. Similarly to before, we have: $ix = 1+x$. This yields the solution: $x = (-1-i)/2, y=(-1+i)/2$.
As far as I can see, these are the only possible solutions... There are three solutions, not nine.
A: Like your earlier question about how to prove the associativity of the group law on elliptic curves, this too is a consequence of general results on intersection theory, esp. (variants of) Bezout's theorem. For computational purposes you can employ various constructive elimination techniques, e.g Gröbner bases, various triangularization methods, etc, which are available in most computer algebra systems
