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It's $(X+Y)^3- (X^2+Y^2)\in \mathbb{C}[X,Y]$ irreducible? I can't apply Eisenstein Criterion.

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    you have to solve $\frac{y^3-y^2}{a}+a(3y-1-a) =3y^2 $2017-01-12

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The brute force method :

See it as a polynomial of degree $3$ of $K[X]$ with coefficients in $K = \mathbb{C}(y)$. If it is not irreducible then $$X^3+y^3+3X^2y+3Xy^2-X^2-y^2=( X +a)(X^2+bX+c)=X^3+bX^2 +cX+aX^2+abX+ac$$

$c = \frac{y^3-y^2}{a}$, $b = 3y-1-a$ $$= X^3+(3y-1-a)X^2 +\frac{y^3-y^2}{a}X+aX^2+a(3y-1-a)X+1$$


I made a mistake, you have to solve $\frac{y^3-y^2}{a}+a(3y-1-a) =3y^2 $ and show the solution $a \not\in \mathbb{C}(y)$