Question

(\frac{sqrt3{9}}{sqrt3{4}})

Explanation:

Step1: Use the property of radicals $\frac{\sqrt[3]{a}}{\sqrt[3]{b}}=\sqrt[3]{\frac{a}{b}}$

$\frac{\sqrt[3]{9}}{\sqrt[3]{4}}=\sqrt[3]{\frac{9}{4}}$

Step2: Simplify the fraction inside the cube root (or leave as is, but we can rationalize or present in simplified radical form)

We can also rewrite $\frac{9}{4}$ as is, and the cube root of a fraction is the fraction of the cube roots. Alternatively, we can multiply numerator and denominator inside the cube root by $4^2$ to rationalize the denominator:
$\sqrt[3]{\frac{9\times4^2}{4\times4^2}}=\sqrt[3]{\frac{9\times16}{4^3}}=\frac{\sqrt[3]{144}}{4}$ (but this is more complicated, the first form is simpler)
Wait, actually, the initial step is correct as $\frac{\sqrt[3]{9}}{\sqrt[3]{4}}=\sqrt[3]{\frac{9}{4}}$ or we can write it as $\sqrt[3]{\frac{9}{4}}$ which can also be expressed as $\frac{\sqrt[3]{36}}{2}$? Wait, no, let's check:

Wait, $9 = 3^2$, $4 = 2^2$, but maybe better to use the rule $\frac{\sqrt[3]{a}}{\sqrt[3]{b}}=\sqrt[3]{\frac{a}{b}}$. So $\frac{\sqrt[3]{9}}{\sqrt[3]{4}}=\sqrt[3]{\frac{9}{4}}$. Alternatively, if we want to rationalize the denominator, multiply numerator and denominator by $\sqrt[3]{4^2}$:

$\frac{\sqrt[3]{9}\times\sqrt[3]{4^2}}{\sqrt[3]{4}\times\sqrt[3]{4^2}}=\frac{\sqrt[3]{9\times16}}{\sqrt[3]{4^3}}=\frac{\sqrt[3]{144}}{4}$. But maybe the problem just wants to simplify the fraction of cube roots, so the first step is to apply the radical division rule.

Wait, the original expression is $\frac{\sqrt[3]{9}}{\sqrt[3]{4}}$. Using the property $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$ for real numbers $a,b$ (with $b
eq0$ and $n$ odd, which 3 is), so we can write it as $\sqrt[3]{\frac{9}{4}}$. Alternatively, we can factor 9 and 4: 9=3², 4=2², but there's no common factors, so the simplified form is $\sqrt[3]{\frac{9}{4}}$ or $\frac{\sqrt[3]{36}}{2}$? Wait, no, $\sqrt[3]{9\times4}=\sqrt[3]{36}$, wait no, 9×4=36, but we have 9/4. Wait, I made a mistake earlier. Let's do it correctly:

$\frac{\sqrt[3]{9}}{\sqrt[3]{4}}=\sqrt[3]{\frac{9}{4}}$. To rationalize the denominator, we multiply numerator and denominator by $\sqrt[3]{4^2}$ (since $4\times4^2=4^3$):

Numerator: $\sqrt[3]{9}\times\sqrt[3]{16}=\sqrt[3]{9\times16}=\sqrt[3]{144}$

Denominator: $\sqrt[3]{4}\times\sqrt[3]{16}=\sqrt[3]{64}=4$

So $\frac{\sqrt[3]{144}}{4}$ can be simplified further? 144=12×12=12², no, 144=2^4×3^2, so $\sqrt[3]{144}=\sqrt[3]{8\times18}=2\sqrt[3]{18}$, so $\frac{2\sqrt[3]{18}}{4}=\frac{\sqrt[3]{18}}{2}$. Wait, let's check:

9×16=144, 144÷8=18, so $\sqrt[3]{144}=\sqrt[3]{8\times18}=2\sqrt[3]{18}$, then $\frac{2\sqrt[3]{18}}{4}=\frac{\sqrt[3]{18}}{2}$. Alternatively, 9/4= (3²)/(2²), but cube root of that is (3²)^(1/3)/(2²)^(1/3)=3^(2/3)/2^(2/3)=(3/2)^(2/3)=∛((3/2)²)=∛(9/4), which is the same as before.

But maybe the problem just wants to simplify the fraction of cube roots, so the key step is using the property of radicals for division. So the simplified form is $\sqrt[3]{\frac{9}{4}}$ or $\frac{\sqrt[3]{36}}{2}$? Wait, no, 9×4=36, so $\sqrt[3]{9/4}=\sqrt[3]{9}/\sqrt[3]{4}$, which is the original form. Wait, maybe I overcomplicated. The correct simplification using the radical division property is $\frac{\sqrt[3]{9}}{\sqrt[3]{4}}=\sqrt[3]{\frac{9}{4}}$, which can also be written as $\frac{\sqrt[3]{36}}{2}$? Wait, no, let's compute 9×4=36, so $\sqrt[3]{9/4}=\sqrt[3]{(9×4)/4²}=\sqrt[3]{36}/\sqrt[3]{16}$? No, that's not helpful.

Wait, the initial expression is $\frac{\sqrt[3]{9}}{\sqrt[3]{4}}$. Let's compute the numerical value to check:

$\sqrt[3]{9}\approx2.08$, $\sqrt[3]{4}\approx1.587$, so 2.08/1.587≈1.31.

$\sqrt[3]{9/4}=\sqrt[3]{2.25}\approx1.31$, which matches.

$\frac{\sqrt[3]{36}}{2}\approx3.3019/2≈1.65$, which is wrong. Wait, I made a mistake earlier. 9×4=36, but when rationalizing, we need to multiply numerator and denominator by $\sqrt[3]{4^2}$, so numerator is $\sqrt[3]{9}\times\sqrt[3]{4^2}=\sqrt[3]{9\times16}=\sqrt[3]{144}\approx5.241$, denominator is 4, so 5.241/4≈1.31, which matches. And $\sqrt[3]{144}=\sqrt[3]{8\times18}=2\sqrt[3]{18}\approx2\times2.6207=5.241$, so $\frac{2\sqrt[3]{18}}{4}=\frac{\sqrt[3]{18}}{2}\approx2.6207/2≈1.31$, which is correct. So $\frac{\sqrt[3]{9}}{\sqrt[3]{4}}=\frac{\sqrt[3]{18}}{2}$ (since 144=8×18, so $\sqrt[3]{144}=2\sqrt[3]{18}$, then divide by 4 gives $\sqrt[3]{18}/2$).

But maybe the problem just wants to apply the radical division rule, so the simplified form is $\sqrt[3]{\frac{9}{4}}$ or $\frac{\sqrt[3]{36}}{2}$? Wait, no, 9/4 is 2.25, cube root of that is approximately 1.31, and $\sqrt[3]{36}/2$ is approximately 3.3019/2≈1.65, which is wrong. Wait, I see my mistake: 9×4=36, but when we have $\frac{\sqrt[3]{9}}{\sqrt[3]{4}}$, multiplying numerator and denominator by $\sqrt[3]{4^2}$ gives $\frac{\sqrt[3]{9\times4^2}}{\sqrt[3]{4\times4^2}}=\frac{\sqrt[3]{9\times16}}{\sqrt[3]{4^3}}=\frac{\sqrt[3]{144}}{4}$, and 144=8×18, so $\sqrt[3]{144}=2\sqrt[3]{18}$, so $\frac{2\sqrt[3]{18}}{4}=\frac{\sqrt[3]{18}}{2}$, which is correct. And $\sqrt[3]{18}\approx2.6207$, so 2.6207/2≈1.31, which matches the decimal approximation of $\sqrt[3]{9/4}$.

Alternatively, maybe the problem is to rationalize the denominator, so the steps are:

1. Apply $\frac{\sqrt[3]{a}}{\sqrt[3]{b}}=\sqrt[3]{\frac{a}{b}}$: $\frac{\sqrt[3]{9}}{\sqrt[3]{4}}=\sqrt[3]{\frac{9}{4}}$

2. Rationalize the denominator by multiplying numerator and denominator inside the cube root by $4^2$:

$\sqrt[3]{\frac{9\times4^2}{4\times4^2}}=\sqrt[3]{\frac{9\times16}{4^3}}=\frac{\sqrt[3]{144}}{4}$

3. Simplify $\sqrt[3]{144}$ as $2\sqrt[3]{18}$ (since 144=8×18, and $\sqrt[3]{8}=2$):

$\frac{2\sqrt[3]{18}}{4}=\frac{\sqrt[3]{18}}{2}$

So the simplified form is $\frac{\sqrt[3]{18}}{2}$ or $\sqrt[3]{\frac{9}{4}}$ or $\frac{\sqrt[3]{144}}{4}$. Depending on the requirement, but the most simplified radical form (with a rational denominator) is $\frac{\sqrt[3]{18}}{2}$.

Answer:

$\boxed{\dfrac{\sqrt[3]{18}}{2}}$ (or $\boxed{\sqrt[3]{\dfrac{9}{4}}}$ depending on the preferred form)