# How do you simplify (\frac { ( 4s ^ { 3} t ^ { - 2} ) ( 2t - 3u ^ { 5} ) } { 8u ^ { 7} } ) ^ { 2}?

Jul 21, 2017

${\left(\frac{\left(4 {s}^{3} {t}^{- 2}\right) \left(2 t - 3 {u}^{5}\right)}{8 {u}^{7}}\right)}^{2} = \frac{{s}^{6} {\left(2 t - 3 {u}^{5}\right)}^{2}}{4 {t}^{4} {u}^{14}}$

#### Explanation:

${\left(\frac{\left(4 {s}^{3} {t}^{- 2}\right) \left(2 t - 3 {u}^{5}\right)}{8 {u}^{7}}\right)}^{2}$

= $\left(\frac{{4}^{2} {\left({s}^{3}\right)}^{2} {\left(\frac{1}{t} ^ 2\right)}^{2} {\left(2 t - 3 {u}^{5}\right)}^{2}}{{8}^{2} {\left({u}^{7}\right)}^{2}}\right)$

= $\frac{16 {s}^{3 \times 2} \left({1}^{2} / {t}^{2 \times 2}\right) {\left(2 t - 3 {u}^{5}\right)}^{2}}{64 {u}^{7 \times 2}}$

= $\frac{\cancel{16} {s}^{6} \left(\frac{1}{t} ^ 4\right) {\left(2 t - 3 {u}^{5}\right)}^{2}}{\cancel{16} \times 4 {u}^{14}}$

= $\frac{{s}^{6} {\left(2 t - 3 {u}^{5}\right)}^{2}}{4 {t}^{4} {u}^{14}}$

Jul 28, 2017

$\frac{{s}^{6} {\left(2 t - 3 {u}^{5}\right)}^{2}}{4 {t}^{4} {u}^{14}}$

#### Explanation:

${\left(\frac{\left(4 {s}^{3} {t}^{\textcolor{b l u e}{- 2}}\right) \left(2 t - 3 {u}^{5}\right)}{8 {u}^{7}}\right)}^{2}$

Simplify inside the bracket first:

$= {\left(\frac{\cancel{4} {s}^{3} \left(2 t - 3 {u}^{5}\right)}{{\cancel{8}}^{2} {t}^{\textcolor{b l u e}{2}} {u}^{7}}\right)}^{2}$

Square each factor in the bracket:

$\frac{{s}^{6} {\left(2 t - 3 {u}^{5}\right)}^{2}}{{2}^{2} {t}^{4} {u}^{14}}$

We could also square the binomial but there seems little point.
We would not be able to cancel any factors (there will be + and -signs.
There will not be like terms nor a common factor