Question

use the quotient rule to simplify. assume that all variables represent positive real numbers.
\sqrt3{\frac{11n}{81m^{12}}}
\sqrt3{\frac{11n}{81m^{12}}} = \square
(type an exact answer, using radicals as needed. simplify your answer.)

Explanation:

Step1: Recall the quotient rule for radicals

The quotient rule for cube roots states that \(\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\) for \(a,b\) real numbers and \(b
eq0\). So we can rewrite \(\sqrt[3]{\frac{11n}{81m^{12}}}\) as \(\frac{\sqrt[3]{11n}}{\sqrt[3]{81m^{12}}}\).

Step2: Simplify the denominator

First, factor \(81\) and \(m^{12}\). We know that \(81 = 27\times3=3^{3}\times3\) and \(m^{12}=(m^{4})^{3}\) (since \((a^{m})^{n}=a^{mn}\), so \((m^{4})^{3}=m^{12}\)). Then \(\sqrt[3]{81m^{12}}=\sqrt[3]{3^{3}\times 3\times(m^{4})^{3}}\). Using the product rule for radicals \(\sqrt[3]{abc}=\sqrt[3]{a}\times\sqrt[3]{b}\times\sqrt[3]{c}\), we get \(\sqrt[3]{3^{3}}\times\sqrt[3]{3}\times\sqrt[3]{(m^{4})^{3}}\). Since \(\sqrt[3]{3^{3}} = 3\) and \(\sqrt[3]{(m^{4})^{3}}=m^{4}\), this simplifies to \(3m^{4}\sqrt[3]{3}\).

Step3: Rationalize the denominator (optional, but to simplify the fraction)

We have \(\frac{\sqrt[3]{11n}}{3m^{4}\sqrt[3]{3}}\). To rationalize the denominator, multiply the numerator and denominator by \(\sqrt[3]{9}\) (because \(\sqrt[3]{3}\times\sqrt[3]{9}=\sqrt[3]{3\times9}=\sqrt[3]{27} = 3\)). So:

\[

$$\begin{align*} \frac{\sqrt[3]{11n}\times\sqrt[3]{9}}{3m^{4}\sqrt[3]{3}\times\sqrt[3]{9}}&=\frac{\sqrt[3]{99n}}{3m^{4}\times3}\\ &=\frac{\sqrt[3]{99n}}{9m^{4}} \end{align*}$$

\]

Wait, let's check that again. Wait, \(\sqrt[3]{3}\times\sqrt[3]{9}=\sqrt[3]{3\times9}=\sqrt[3]{27} = 3\), so the denominator becomes \(3m^{4}\times3 = 9m^{4}\) and the numerator is \(\sqrt[3]{11n}\times\sqrt[3]{9}=\sqrt[3]{99n}\) (since \(11\times9 = 99\)). Wait, no, actually, when we multiply \(\sqrt[3]{11n}\times\sqrt[3]{9}=\sqrt[3]{11n\times9}=\sqrt[3]{99n}\). And the denominator: \(\sqrt[3]{81m^{12}}=\sqrt[3]{3^{3}\times3\times(m^{4})^{3}} = 3m^{4}\sqrt[3]{3}\), so going back to \(\frac{\sqrt[3]{11n}}{3m^{4}\sqrt[3]{3}}\), multiply numerator and denominator by \(\sqrt[3]{9}\):

Numerator: \(\sqrt[3]{11n}\times\sqrt[3]{9}=\sqrt[3]{99n}\)

Denominator: \(3m^{4}\times\sqrt[3]{3}\times\sqrt[3]{9}=3m^{4}\times\sqrt[3]{27}=3m^{4}\times3 = 9m^{4}\)

So the simplified form is \(\frac{\sqrt[3]{99n}}{9m^{4}}\)

Wait, another way: Let's re - express the original fraction:

\(\sqrt[3]{\frac{11n}{81m^{12}}}=\sqrt[3]{\frac{11n\times3^{2}}{81m^{12}\times3^{2}}}\) (multiplying numerator and denominator inside the cube root by \(9 = 3^{2}\) to make the denominator a perfect cube). Then the denominator inside the cube root is \(81\times9m^{12}=729m^{12}=(9m^{4})^{3}\) (since \(9^{3}=729\) and \((m^{4})^{3}=m^{12}\)). And the numerator inside the cube root is \(11n\times9 = 99n\). So \(\sqrt[3]{\frac{99n}{(9m^{4})^{3}}}=\frac{\sqrt[3]{99n}}{9m^{4}}\) (because \(\sqrt[3]{\frac{a}{b^{3}}}=\frac{\sqrt[3]{a}}{b}\) for \(b
eq0\)).

Answer:

\(\frac{\sqrt[3]{99n}}{9m^{4}}\)