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

May 17, 2016

$\frac{x + y}{\left({x}^{2} + {y}^{2}\right) \left(x - y\right)}$

#### Explanation:

There are two polynomial equalities that will help us.
$\left(x - a\right) \left(x - a\right) = {x}^{2} - {a}^{2}$ and ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$
Analyzing the numerator ${x}^{4} - {y}^{4} = \left({x}^{2} + {y}^{2}\right) \left({x}^{2} - {y}^{2}\right)$
At the denominator we have
${x}^{4} + 2 {x}^{2} {y}^{2} + {y}^{4} = {\left({x}^{2} + {y}^{2}\right)}^{2}$ and ${x}^{2} - 2 x y + {y}^{2} = {\left(x - y\right)}^{2}$
Putting all together
$\frac{\left({x}^{2} + {y}^{2}\right) \left({x}^{2} - {y}^{2}\right)}{{\left({x}^{2} + {y}^{2}\right)}^{2} {\left(x - y\right)}^{2}} = \frac{x + y}{\left({x}^{2} + {y}^{2}\right) \left(x - y\right)}$

May 17, 2016

$\frac{{x}^{4} - {y}^{4}}{\left({x}^{4} + 2 {x}^{2} {y}^{2} + {y}^{4}\right) \left({x}^{2} - 2 x y + {y}^{2}\right)} = \frac{x + y}{\left({x}^{2} + {y}^{2}\right) \left(x - y\right)}$

#### Explanation:

$\frac{{x}^{4} - {y}^{4}}{\left({x}^{4} + 2 {x}^{2} {y}^{2} + {y}^{4}\right) \left({x}^{2} - 2 x y + {y}^{2}\right)}$

$= \frac{\left({x}^{2} - {y}^{2}\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left({x}^{2} + {y}^{2}\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left({x}^{2} + {y}^{2}\right)}}} \left({x}^{2} + {y}^{2}\right) \left(x - y\right) \left(x - y\right)}$

$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - y\right)}}} \left(x + y\right)}{\left({x}^{2} + {y}^{2}\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - y\right)}}} \left(x - y\right)}$

$= \frac{x + y}{\left({x}^{2} + {y}^{2}\right) \left(x - y\right)}$

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.