# How do you simplify (64a^2b^5)/(35b^2c^3f^4)div(12a^4b^3c)/(70abcf^2)?

Jan 8, 2017

See full explanation below:

#### Explanation:

First step we will take is to rewrite this expression in the form below:

$\frac{\frac{64 {a}^{2} {b}^{5}}{35 {b}^{2} {c}^{3} {f}^{4}}}{\frac{12 {a}^{4} {b}^{3} c}{70 a b c {f}^{2}}}$

Next, we will divide these fractions using the rule:

$\frac{\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}}{\frac{\textcolor{g r e e n}{c}}{\textcolor{p u r p \le}{d}}} = \frac{\textcolor{red}{a} \times \textcolor{p u r p \le}{d}}{\textcolor{b l u e}{b} \times \textcolor{g r e e n}{c}}$

$\frac{\left(64 {a}^{2} {b}^{5}\right) \times \left(70 a b c {f}^{2}\right)}{\left(35 {b}^{2} {c}^{3} {f}^{4}\right) \times \left(12 {a}^{4} {b}^{3} c\right)}$

Then, we can multiply the terms in the numerator and the terms in the denominator using the rules for exponents:

${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$x = {x}^{\textcolor{red}{1}}$

$\frac{4480 {a}^{2 + 1} {b}^{5 + 1} c {f}^{2}}{420 {a}^{4} {b}^{2 + 3} {c}^{3 + 1} {f}^{4}}$

$\frac{4480 {a}^{3} {b}^{6} c {f}^{2}}{420 {a}^{4} {b}^{5} {c}^{4} {f}^{4}}$

We use these rules for exponents to divide this expression:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$

$\frac{4480 {b}^{6 - 5}}{420 {a}^{4 - 3} {c}^{4 - 1} {f}^{4 - 2}}$

$\frac{4480 {b}^{1}}{420 {a}^{1} {c}^{3} {f}^{2}}$

$\frac{4480 b}{420 a {c}^{3} {f}^{2}}$

Now, we can finish by factoring the constants:

$\frac{\left(140 \times 64\right) b}{\left(140 \times 3\right) a {c}^{3} {f}^{2}}$

$\frac{\left(\cancel{140} \times 64\right) b}{\left(\cancel{140} \times 3\right) a {c}^{3} {f}^{2}}$

$\frac{64 b}{3 a {c}^{3} {f}^{2}}$