# Perform indicated operation?

## $\frac{2 p}{4 {p}^{2} - 1} / \frac{6 {p}^{3}}{6 p + 3}$

Jun 20, 2018

$\frac{\frac{2 p}{4 {p}^{2} - 1}}{\frac{6 {p}^{3}}{6 p + 3}}$ = $\frac{1}{{p}^{2} \left(2 p - 1\right)}$

#### Explanation:

$\frac{\frac{2 p}{4 {p}^{2} - 1}}{\frac{6 {p}^{3}}{6 p + 3}}$

This would be the same as multiplying the numerator
$\frac{2 p}{4 {p}^{2} - 1}$
with the inverse of the denominator
$\frac{6 {p}^{3}}{6 p + 3}$, i.e. $\frac{6 p + 3}{6 {p}^{3}}$

We, therefore, get:
$\frac{\frac{2 p}{4 {p}^{2} - 1}}{\frac{6 {p}^{3}}{6 p + 3}}$
= $\left(\frac{2 p}{4 {p}^{2} - 1}\right) \cdot \left(\frac{6 p + 3}{6 {p}^{3}}\right)$ = $\frac{2 p \left(6 p + 3\right)}{6 {p}^{3} \left(4 {p}^{2} - 1\right)}$
= $\frac{3 \left(2 p + 1\right)}{3 {p}^{2} \left(4 {p}^{2} - 1\right)}$ = $\frac{\left(2 p + 1\right)}{{p}^{2} \left(4 {p}^{2} - 1\right)}$

Now we need to remember that (a+b)(a-b)=a^2-b^2
Therefore (2x+1)(2x-1)=(4x^2-1)

We can, therefore, write:
$\frac{\left(2 p + 1\right)}{{p}^{2} \left(2 p + 1\right) \left(2 p - 1\right)} = \frac{1}{{p}^{2} \left(2 p - 1\right)}$