# How do you evaluate \frac { b ^ { \frac { 2} { 2} } } { 4b ^ { 2} } - \frac { 2b ^ { - 3} } { 4b ^ { 2} }?

##### 1 Answer
Dec 16, 2017

$= \setminus \frac{{b}^{4} - 2}{4 {b}^{5}}$

#### Explanation:

$\setminus \frac{{b}^{\setminus \frac{2}{2}}}{4 {b}^{2}} - \setminus \frac{2 {b}^{- 3}}{4 {b}^{2}}$

$= \setminus \frac{{b}^{1}}{4 {b}^{2}} - \setminus \frac{2 {b}^{- 3}}{4 {b}^{2}}$

As the denominators are equal,

$= \frac{{b}^{1} - 2 {b}^{-} 3}{4 {b}^{2}}$

$= \frac{b \left(1 - 2 {b}^{-} 4\right)}{4 {b}^{2}}$

$= \frac{\cancel{b} \left(1 - 2 {b}^{-} 4\right)}{4 {b}^{\cancel{2}}}$

$= \frac{1 - 2 {b}^{-} 4}{4 b}$

Rule ${a}^{-} m = \frac{1}{a} ^ m$

$= \frac{1 - 2 \left(\frac{1}{b} ^ 4\right)}{4 {b}^{1}}$

Make denominators of the numerator fraction equal :

$= \frac{1 \left({b}^{4} / {b}^{4}\right) - 2 \left(\frac{1}{b} ^ 4\right)}{4 {b}^{1}}$

$= \frac{\frac{\left({b}^{4} - 2\right)}{{b}^{4}}}{4 {b}^{1}}$

$= \frac{{b}^{4} - 2}{{b}^{4}} \times \frac{1}{4 {b}^{1}}$

Rule: ${a}^{m} \times {a}^{n} = {a}^{m + n}$

$= \frac{{b}^{4} - 2}{4 {b}^{5}}$