Factorisation

**muks** · November 9th 2008, 06:21 AM

Factorise:

$(xy+1)^4 - 4xy(xy+1)^2 - (x^2 - y^2)^2$

Thank you.

**Shyam** · November 9th 2008, 08:21 AM

Originally Posted by muks

Factorise:

$(xy+1)^4 - 4xy(xy+1)^2 - (x^2 - y^2)^2$

Thank you.

$(xy+1)^4 - 4xy(xy+1)^2 - (x^2 - y^2)^2$

As you see, $(xy+1)^2$ comes twice, so, Complete the square

$=[(xy+1)^2]^2-2(2xy)(xy+1)^2+(2xy)^2-(2xy)^2-(x^2-y^2)^2$

$=[(xy+1)^2-(2xy)]^2-(2xy)^2-(x^2-y^2)^2$

$=[x^2y^2+1+2xy-2xy]^2-4x^2y^2-[(x^2)^2+(y^2)^2-2x^2y^2]$

$=(x^2y^2+1)^2-[4x^2y^2+(x^2)^2+(y^2)^2-2x^2y^2]$

$=(x^2y^2+1)^2-[(x^2)^2+(y^2)^2+2x^2y^2]$

$=(x^2y^2+1)^2-(x^2+y^2)^2$

$=(x^2y^2+1+x^2+y^2)(x^2y^2+1-x^2-y^2)$

**muks** · November 10th 2008, 07:11 AM

I agree with the answer provided, but the answer in the book is $(x+1)(x-1)(y+1)(y-1)(x^2+1)(y^2+1)$

How did they obtain?

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