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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

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The command of Maple $$ factor(-8*x^3*y+8*x*y^3+8*x^2-8*y^2)$$ produces $$-8\, \left( -y+x \right) \left( y+x \right) \left( xy-1 \right) .$$ –  user64494 Jul 26 '13 at 19:47

4 Answers 4

up vote 5 down vote accepted

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.

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First take out $8$ as factor

$8xy^3+8x^2-8x^3y-8y^2=8(xy^3-y^2+x^2-x^3y)$

Now, $xy^3-y^2+x^2-x^3y=y^2(xy-1)-x^2(xy-1)=(xy-1)(y^2-x^2)=(xy-1)(y+x)(y-x)$

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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.

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$$ 8xy^3+8x^2-8x^3y-8y^2= 8xy^3-8x^3y+8x^2-8y^2$$

$$= 8xy(y^2-x^2)-8(y^2-x^2)= (8xy-8)(y^2-x^2) $$

$$ = 8(1-xy)(x^2-y^2). $$

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