Question

How do you simplify the expression `(x^4-y^4) /( (x^4+2x^2y^2+y^4)(x^2-2xy+y^2))`?

Answer 1

`(x+y)/((x^2+y^2)(x-y))`

Explanation:

There are two polynomial equalities that will help us.
`(x-a)(x-a)=x^2-a^2` and `(x + a)^2=x^2+2ax+a^2`
Analyzing the numerator `x^4-y^4 = (x^2+y^2)(x^2-y^2)`
At the denominator we have
`x^4+2x^2y^2+y^4=(x^2+y^2)^2` and `x^2-2xy+y^2=(x-y)^2`
Putting all together
`((x^2+y^2)(x^2-y^2))/((x^2+y^2)^2(x-y)^2)=(x+y)/((x^2+y^2)(x-y))`

Answer 2

`(x^4-y^4)/((x^4+2x^2y^2+y^4)(x^2-2xy+y^2))=(x+y)/((x^2+y^2)(x-y))`

Explanation:

`(x^4-y^4)/((x^4+2x^2y^2+y^4)(x^2-2xy+y^2))`

`=((x^2-y^2)color(red)(cancel(color(black)((x^2+y^2)))))/(color(red)(cancel(color(black)((x^2+y^2))))(x^2+y^2)(x-y)(x-y))`

`=(color(red)(cancel(color(black)((x-y))))(x+y))/((x^2+y^2)color(red)(cancel(color(black)((x-y))))(x-y))`

`=(x+y)/((x^2+y^2)(x-y))`

Note that we do not have to specify any exclusions as the values we have cancelled out from the numerator and denominator exist in the denominator of the simplified expression.