How would one go about solving a complicated polynomials? (Not necessarily complex one's, though) Say I have an equation such as:
$$
x^2 + 3x^4y^2 + 3y^4x^2 + y^6 = 5x^2y^2
$$
How would one go about solving such an equation? Where $y$ could be isolated in terms of $x$. I.e, $y = $ formula involving $x$.
 A: If you want find the solutions for $y$ you have to ordinate the polynomial with the $y$ powers and in your case we have:
$$
y^6+3x^2y^4+y^2(3x^4-5x^2)+x^2=0
$$
In general this is a polynomial of degree $n$ (in you case $n=6$) and, for $n>4$ we have not general formulas that gives a solution by means of radicals ( Abel-Ruffini theorem). In this case we can put $y^2=t$ and the equation becomes
$$
t^3+3x^2t^2+(3x^4-5x^2)t+x^2=0
$$
That is a third degree equation with a ''parameter'' $x$. So, if we can solve such equation we can find a formula that gives $y^2=F(x)$. 
Depending if you want for $y$ real or complex numbers you have to discuss such solution to find the good values of $x$ .
I don't see a simple way to solve it without using the Cardano formula.(and the discussion of solutions is not easy at all).
A: Maybe a slightly easier approach is to complete the cube first as
$$
x^2+(x^2+y^2)^3-x^6=5x^2y^2
$$
and introduce the variable $z=x^2+y^2$ to get
$$
0=x^2+z^3-x^6-5x^2(z-x^2)=z^3-5x^2z-x^6+5x^4+x^2
$$
which is a depressed cubic (without a quadratic term) in $z$.
One can further divide the depressed cubic by $x^3$ (assuming $x\ne 0$) and make the change $t=\frac{z}{x}$ to get
$$
f(x,t)=t^3-5t-x^3+5x+\frac{1}{x}=0
$$
P.S. It is easy to show that the only real solution to the original equation is $x=y=0$. For example, one can minimize the function $f(x,t)$ for $x>0$, $t\ge x$ and see that the minimum is positive.
