# How do you simplify (27a^-3b^12)^(1/3) / (16a^-8b^12) ^ (1/2)?

Jun 6, 2016

You must remember the exponent-radical rule ${x}^{\frac{1}{n}} = \sqrt[n]{x}$

#### Explanation:

Therefore,

$\frac{\sqrt[3]{27} \times {\left({a}^{-} 3 \times {b}^{12}\right)}^{\frac{1}{3}}}{\sqrt{16} \times {\left({a}^{-} 8 \times {b}^{12}\right)}^{\frac{1}{2}}}$

For the expressions in parentheses, you must calculate using the power of exponents rule, or ${\left({a}^{n}\right)}^{m} = {a}^{n \times m}$

$= \frac{3 \times {a}^{-} 1 \times {b}^{4}}{4 \times {a}^{-} 4 \times {b}^{6}}$

However, we need to simplify further and write without negative exponents. This can all be done using the quotient rule ${a}^{n} / {a}^{m} = {a}^{n - m}$

As a shortcut to not have to use the negative exponent rule ${a}^{-} n = \frac{1}{a} ^ n$, we must apply the quotient rule from the largest exponent. For example, in ${x}^{2} / {x}^{4}$, you would make ${x}^{4}$ as n and ${x}^{2}$ as m, and then you would do your subtraction. You would get $\frac{1}{x} ^ 2$ in this problem, which is without negative exponents, and is what we want.

$= \frac{3 \times {a}^{- 1 - \left(- 4\right)}}{4 \times {b}^{6 - 4}}$

$= \frac{3 {a}^{3}}{4 {b}^{2}}$

Hopefully this helps!