# How do you simplify ((x^2-y^2)(x^2+xy+y^2))/((x^3-y^3)(x^2+2xy+y^2))?

Apr 15, 2018

It simplifies to $\frac{1}{x + y}$.

#### Explanation:

First, factor the bottom right and top left polynomials using the special binomial factoring cases:

$\textcolor{w h i t e}{=} \frac{\textcolor{g r e e n}{\left({x}^{2} - {y}^{2}\right)} \left({x}^{2} + x y + {y}^{2}\right)}{\left({x}^{3} - {y}^{3}\right) \textcolor{b l u e}{\left({x}^{2} + 2 x y + {y}^{2}\right)}}$

$= \frac{\textcolor{g r e e n}{\left(x - y\right) \left(x + y\right)} \left({x}^{2} + x y + {y}^{2}\right)}{\left({x}^{3} - {y}^{3}\right) \textcolor{b l u e}{\left(x + y\right) \left(x + y\right)}}$

Cancel the common factor:

$= \frac{\textcolor{g r e e n}{\left(x - y\right) \textcolor{red}{\cancel{\textcolor{g r e e n}{\left(x + y\right)}}}} \left({x}^{2} + x y + {y}^{2}\right)}{\left({x}^{3} - {y}^{3}\right) \textcolor{b l u e}{\left(x + y\right) \textcolor{red}{\cancel{\textcolor{b l u e}{\left(x + y\right)}}}}}$

$= \frac{\textcolor{g r e e n}{\left(x - y\right)} \left({x}^{2} + x y + {y}^{2}\right)}{\left({x}^{3} - {y}^{3}\right) \textcolor{b l u e}{\left(x + y\right)}}$

Next, use the difference of cubes product to factor the bottom left polynomial:

$= \frac{\textcolor{g r e e n}{\left(x - y\right)} \left({x}^{2} + x y + {y}^{2}\right)}{\textcolor{m a \ge n t a}{\left({x}^{3} - {y}^{3}\right)} \textcolor{b l u e}{\left(x + y\right)}}$

$= \frac{\textcolor{g r e e n}{\left(x - y\right)} \left({x}^{2} + x y + {y}^{2}\right)}{\textcolor{m a \ge n t a}{\left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)} \textcolor{b l u e}{\left(x + y\right)}}$

Cancel the common factors again:

$= \frac{\textcolor{red}{\cancel{\textcolor{g r e e n}{\left(x - y\right)}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\left({x}^{2} + x y + {y}^{2}\right)}}}}{\textcolor{m a \ge n t a}{\textcolor{red}{\cancel{\textcolor{m a \ge n t a}{\left(x - y\right)}}} \textcolor{red}{\cancel{\textcolor{m a \ge n t a}{\left({x}^{2} + x y + {y}^{2}\right)}}}} \textcolor{b l u e}{\left(x + y\right)}}$

$= \frac{1}{\textcolor{b l u e}{x + y}}$

That's as simplified as it gets. Hope this helped!

Apr 15, 2018

$\frac{1}{x + y}$

#### Explanation:

I'll use the following formulas:

• $\textcolor{b l u e}{{x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)}$
• $\textcolor{p u r p \le}{{x}^{3} - {y}^{3} = \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)}$
• $\textcolor{g r e e n}{{\left(x + y\right)}^{2} = {x}^{2} + 2 x y + {y}^{2}}$

( color(blue)((x^2 - y^2 )) (x^2 + xy + y^2)) / (color(purple)((x^3 - y^3)) color(green)((x^2 + 2xy + y^2))

$= \frac{\textcolor{b l u e}{\left(x + y\right) \left(x - y\right)} \left({x}^{2} + x y + {y}^{2}\right)}{\textcolor{p u r p \le}{\left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)} \textcolor{g r e e n}{{\left(x + y\right)}^{2}}}$

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

$= \frac{x + y}{x + y} ^ 2$

$= \frac{1}{x + y}$