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Does anyone know how to factorize the following expression:

$$4x^4+12x^{10/3} y^{2/3}+33x^{8/3} y^{4/3}+46x^2 y^2+33x^{4/3} y^{8/3}+12x^{2/3} y^{10/3}+4 y^4$$ ?

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I removed some tags, changed one, added one, changed title and formatting of question. – Aryabhata Dec 29 '10 at 6:11
@Mo: You know how it is, some expressions are just too damn big to be placed in question titles... :D – J. M. Dec 29 '10 at 6:18
@J.M: Yup ... :-) – Aryabhata Dec 29 '10 at 6:19
up vote 8 down vote accepted

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).

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