How to factor $8xy^3+8x^2-8x^3y-8y^2$ How can I factor $8xy^3+8x^2-8x^3y-8y^2$ or the different form $2x(4y^3+4x)-2y(4x^3+4y)$
Is there any general methods that work?
A possible solution should be $8(x^2-y^2)(1-xy)$ But please do not start from here as in the general case I will not know the answer...
Thanks!
Alexander
 A: For polynomials with four terms, try grouping them:
$$8(\underbrace{xy^3-x^3y}+\underbrace{x^2-y^2})=8[-xy(x^2-y^2)+x^2-y^2]=8(1-xy)(x^2-y^2)=\boxed{8(1-xy)(x-y)(x+y)}$$
I factored out a $-xy$ from the first group to make the quotient look like the second group.  This allowed me to then factor out $x^2-y^2$ from both terms in the square brackets.
A: First take out $8$ as factor
$8xy^3+8x^2-8x^3y-8y^2=8(xy^3-y^2+x^2-x^3y)$
Now,
$xy^3-y^2+x^2-x^3y=y^2(xy-1)-x^2(xy-1)=(xy-1)(y^2-x^2)=(xy-1)(y+x)(y-x)$
A: I don't think there is a general method. 
However spotting obvious factors - like $8$ - and grouping factors of the same degree together so $x^2$ and $y^2$ have degree $2$ and $xy^3$ and $x^3y$ have degree 4 - is a useful step to have in mind.
A: $$ 8xy^3+8x^2-8x^3y-8y^2=  8xy^3-8x^3y+8x^2-8y^2$$
$$= 8xy(y^2-x^2)-8(y^2-x^2)= (8xy-8)(y^2-x^2) $$
$$ = 8(1-xy)(x^2-y^2). $$
A: Since $f(x,y)=8xy^3+8x^2-8x^3y-8y^2$ is zero for $y=x,-x$, it appears that
$$
f(x,y)=8(x-y)(x+y)g(x,y).
$$
Now it is easy to find that $g(x,y)=1-xy$.
