# How do you divide \frac{a^2+2ab+b^2}{ab^2-a^2b} \-: (a+b)?

Dec 25, 2014

We can use the rule about division of rational expressions where you can change the division in a multiplication by flipping the second fraction.
In our case you have (remember that $\left(a + b\right)$ can be written as a fraction of $\frac{a + b}{1}$):

$\frac{{a}^{2} + 2 a b + {b}^{2}}{a {b}^{2} - {a}^{2} b} \div \frac{a + b}{1} = \frac{{a}^{2} + 2 a b + {b}^{2}}{a {b}^{2} - {a}^{2} b} \times \frac{1}{a + b} =$

You can also rearrange the nominator and denominator of the first fraction as:

${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$ and:
$a {b}^{2} - {a}^{2} b = a b \left(b - a\right)$

So, finally:

$\frac{{a}^{2} + 2 a b + {b}^{2}}{a {b}^{2} - {a}^{2} b} \times \frac{1}{a + b} = \frac{{\left(a + b\right)}^{2}}{a b \left(b - a\right)} \times \frac{1}{a + b} =$
$= \frac{a + b}{a b \left(b - a\right)}$