How to factorize $4x^4+12x^{10/3} y^{2/3}+ \dots $? Does anyone know how to factorize the following expression: 
$$4x^4+12x^{10/3} y^{2/3}+33x^{8/3} y^{4/3}+46x^2 y^2+33x^{4/3} y^{8/3}+12x^{2/3} y^{10/3}+4 y^4$$ 
?

It is possible, but likely messy.
First set $\displaystyle z = \left(\frac{x}{y}\right)^{2/3}$ and divide your expression by $\displaystyle y^4$.
We get
$$4z^6 + 12 z^5 + 33 z^4 + 46 z^3 + 33 z^2 + 12 z + 4$$
Now divide this by $\displaystyle z^3$ and set $\displaystyle t = z + 1/z$.
We get
$$4(z^3 + 1/z^3) + 12(z^2 + 1/z^2) + 33(z+1/z) + 46$$
$$ = 4(t^3 - 3t) + 12(t^2 - 2) + 33t + 46$$
$$ = 4t^3 + 12t^2 + 21t + 22$$
Which is a messy cubic, according to Wolfram Alpha. 
(Note it is always possible to factorize a cubic in "closed form", because of Cardano's method of finding the roots).
