Question

1. \\(\frac{sqrt3{54}}{sqrt3{2}}\\)\
2. \\(\frac{sqrt3{350}}{sqrt3{500}}\\)\
3. \\(\frac{2sqrt{3}}{sqrt{8}}\\)

Explanation:

Problem 23: $\boldsymbol{\frac{\sqrt[3]{54}}{\sqrt[3]{2}}}$

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

$\frac{\sqrt[3]{54}}{\sqrt[3]{2}}=\sqrt[3]{\frac{54}{2}}$

Step2: Simplify the fraction inside the cube root

$\sqrt[3]{\frac{54}{2}}=\sqrt[3]{27}$

Step3: Evaluate the cube root of 27

Since $3^3 = 27$, $\sqrt[3]{27}=3$

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

$\frac{\sqrt[3]{350}}{\sqrt[3]{500}}=\sqrt[3]{\frac{350}{500}}$

Step2: Simplify the fraction inside the cube root

$\frac{350}{500}=\frac{7}{10}$, so we have $\sqrt[3]{\frac{7}{10}}$ or we can rationalize (but usually leaving as $\sqrt[3]{\frac{7}{10}}$ or simplifying the fraction first)
Wait, actually, let's factor numerator and denominator:
$350 = 5\times70=5\times5\times14 = 5^2\times14$
$500=5\times100 = 5\times5\times20=5^2\times20=5^2\times4\times5 = 5^3\times4$ Wait, no, $500 = 5\times100=5\times5\times20 = 5^2\times20=5^2\times4\times5=5^3\times4$? Wait, $5^3=125$, $125\times4 = 500$, yes. And $350=5^2\times14$. So $\frac{350}{500}=\frac{5^2\times14}{5^3\times4}=\frac{14}{5\times4}=\frac{14}{20}=\frac{7}{10}$. So $\sqrt[3]{\frac{7}{10}}$ can be written as $\frac{\sqrt[3]{7}}{\sqrt[3]{10}}$ or rationalized by multiplying numerator and denominator inside the cube root by $10^2$:
$\sqrt[3]{\frac{7\times100}{10\times100}}=\sqrt[3]{\frac{700}{1000}}=\frac{\sqrt[3]{700}}{10}$, but maybe the simplest form is $\sqrt[3]{\frac{7}{10}}$ or we can check if we made a mistake. Wait, original problem: $\frac{\sqrt[3]{350}}{\sqrt[3]{500}}$. Let's do it again:

$\frac{\sqrt[3]{350}}{\sqrt[3]{500}}=\sqrt[3]{\frac{350}{500}}=\sqrt[3]{\frac{7}{10}}$ (after dividing numerator and denominator by 50). Alternatively, if we want to simplify the cube root:

But maybe the problem expects simplifying the fraction first. So $\frac{350}{500}=\frac{7}{10}$, so $\sqrt[3]{\frac{7}{10}}$ is the simplified form, or we can write it as $\frac{\sqrt[3]{700}}{10}$ (by rationalizing: $\sqrt[3]{\frac{7}{10}}=\frac{\sqrt[3]{7\times10^2}}{\sqrt[3]{10\times10^2}}=\frac{\sqrt[3]{700}}{10}$). But maybe the answer is $\sqrt[3]{\frac{7}{10}}$ or $\frac{\sqrt[3]{700}}{10}$. Wait, let's check the calculation again.

Wait, $350 = 5\times70 = 5\times5\times14 = 5^2\times14$; $500 = 5\times100 = 5\times5\times20 = 5^2\times20 = 5^2\times4\times5 = 5^3\times4$. So $\frac{350}{500}=\frac{5^2\times14}{5^3\times4}=\frac{14}{5\times4}=\frac{14}{20}=\frac{7}{10}$. So $\sqrt[3]{\frac{7}{10}}$ is correct. Alternatively, if we want to write it as a single cube root, it's $\sqrt[3]{\frac{7}{10}}$.

Step1: Simplify $\sqrt{8}$

$\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}$

Step2: Substitute back into the expression

$\frac{2\sqrt{3}}{2\sqrt{2}}$

Step3: Cancel the common factor of 2

$\frac{\sqrt{3}}{\sqrt{2}}$

Step4: Rationalize the denominator by multiplying numerator and denominator by $\sqrt{2}$

$\frac{\sqrt{3}\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{\sqrt{6}}{2}$

Answer:

$3$

Problem 24: $\boldsymbol{\frac{\sqrt[3]{350}}{\sqrt[3]{500}}}$