Question

use the quotient rule to simplify. assume that all variables represent positive real numbers.
\sqrt3{\frac{7x}{81y^{12}}}
\sqrt3{\frac{7x}{81y^{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 $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$ (where $b
eq0$). So we can apply this to $\sqrt[3]{\frac{7x}{81y^{12}}}$ to get $\frac{\sqrt[3]{7x}}{\sqrt[3]{81y^{12}}}$.

Step2: Simplify the denominator

First, factor $81$: $81 = 27\times3=3^{3}\times3$. And $y^{12}=(y^{4})^{3}$. So $\sqrt[3]{81y^{12}}=\sqrt[3]{3^{3}\times3\times(y^{4})^{3}}$. Using the product rule for radicals $\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}$ (for real numbers $a,b$), we have $\sqrt[3]{3^{3}\times3\times(y^{4})^{3}}=\sqrt[3]{3^{3}}\cdot\sqrt[3]{(y^{4})^{3}}\cdot\sqrt[3]{3}=3y^{4}\sqrt[3]{3}$.

Step3: Rationalize the denominator (or simplify the fraction)

Now we have $\frac{\sqrt[3]{7x}}{3y^{4}\sqrt[3]{3}}$. To rationalize the denominator, we multiply the numerator and denominator by $\sqrt[3]{3^{2}}$ (since we need to make the radicand in the denominator a perfect cube; $3\times3^{2}=3^{3}$). So:

$$
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$$

Wait, alternatively, we can rewrite the original fraction before rationalizing by expressing $81$ as $3^4$? Wait, no, earlier factoring was $81 = 3^4=3^{3}\times3$. Let's re - examine step 2. Wait, $\sqrt[3]{81y^{12}}=\sqrt[3]{3^{4}y^{12}}=\sqrt[3]{3^{3}\times3\times(y^{4})^{3}}=3y^{4}\sqrt[3]{3}$, that part is correct. Then, when we have $\frac{\sqrt[3]{7x}}{3y^{4}\sqrt[3]{3}}$, multiply numerator and denominator by $\sqrt[3]{3^{2}}$:

Numerator: $\sqrt[3]{7x}\times\sqrt[3]{9}=\sqrt[3]{63x}$

Denominator: $3y^{4}\times\sqrt[3]{3}\times\sqrt[3]{9}=3y^{4}\times\sqrt[3]{27}=3y^{4}\times3 = 9y^{4}$

So the simplified form is $\frac{\sqrt[3]{63x}}{9y^{4}}$? Wait, no, wait the original fraction is $\sqrt[3]{\frac{7x}{81y^{12}}}=\frac{\sqrt[3]{7x}}{\sqrt[3]{81y^{12}}}$. Let's factor $81$ as $3^4$ and $y^{12}=(y^4)^3$. So $\sqrt[3]{81y^{12}}=\sqrt[3]{3^4y^{12}}=\sqrt[3]{3^3\times3\times(y^4)^3}=3y^4\sqrt[3]{3}$. Then, to make the denominator a perfect cube in the radical, we can also rewrite the original expression as:

$\sqrt[3]{\frac{7x}{81y^{12}}}=\sqrt[3]{\frac{7x\times3^{2}}{81y^{12}\times3^{2}}}=\sqrt[3]{\frac{63x}{81\times9y^{12}}}=\sqrt[3]{\frac{63x}{729y^{12}}}$. Since $729 = 9^{3}$ and $y^{12}=(y^{4})^{3}$, then $\sqrt[3]{\frac{63x}{729y^{12}}}=\frac{\sqrt[3]{63x}}{\sqrt[3]{729y^{12}}}=\frac{\sqrt[3]{63x}}{9y^{4}}$.

Yes, that's a better way. By multiplying numerator and denominator inside the cube root by $3^{2}$ (to make the denominator's radicand a perfect cube: $81\times9 = 729=9^{3}$ and $y^{12}=(y^{4})^{3}$), so we have:

$\sqrt[3]{\frac{7x\times9}{81y^{12}\times9}}=\sqrt[3]{\frac{63x}{729y^{12}}}=\frac{\sqrt[3]{63x}}{\sqrt[3]{729y^{12}}}=\frac{\sqrt[3]{63x}}{9y^{4}}$

Answer:

$\frac{\sqrt[3]{63x}}{9y^{4}}$