Question

1. \sqrt3{\frac{500xy^2}{x^4y^{-8}}}

Explanation:

Step1: Simplify the fraction inside the cube root

First, use the quotient rule for exponents \( \frac{a^m}{a^n}=a^{m - n} \) to simplify the variables. For \( x \): \( x^{1-4}=x^{-3}=\frac{1}{x^3} \), for \( y \): \( y^{2-(-8)} = y^{10} \). So the expression inside the cube root becomes \( \frac{500y^{10}}{x^3} \).
\[
\sqrt[3]{\frac{500xy^{2}}{x^{4}y^{-8}}}=\sqrt[3]{\frac{500y^{2 - (-8)}}{x^{4 - 1}}}=\sqrt[3]{\frac{500y^{10}}{x^{3}}}
\]

Step2: Factor out perfect cubes

We know that \( 500 = 100\times5=4\times25\times5 = 2^2\times5^3 \), \( y^{10}=y^{9 + 1}=y^{3\times3+1}=(y^3)^3\times y \), and \( x^3=(x)^3 \). So we can rewrite the expression as:
\[
\sqrt[3]{\frac{2^2\times5^3\times(y^3)^3\times y}{x^3}}
\]

Step3: Apply the cube root property

The cube root of a product is the product of the cube roots, and \( \sqrt[3]{a^3}=a \). So we have:
\[
\frac{\sqrt[3]{5^3}\times\sqrt[3]{(y^3)^3}\times\sqrt[3]{2^2\times5^0\times y}}{\sqrt[3]{x^3}}=\frac{5y\sqrt[3]{4y}}{x}
\]
Wait, wait, let's re - check step 2. Wait, \( 500 = 5\times100=5\times4\times25 = 5\times2^{2}\times5^{2}=2^{2}\times5^{3} \). Yes, that's correct. Then \( \sqrt[3]{\frac{500y^{10}}{x^{3}}}=\sqrt[3]{\frac{5^{3}\times4\times y^{9}\times y}{x^{3}}}=\frac{\sqrt[3]{5^{3}}\times\sqrt[3]{y^{9}}\times\sqrt[3]{4y}}{\sqrt[3]{x^{3}}} \)
Since \( \sqrt[3]{5^{3}} = 5 \), \( \sqrt[3]{y^{9}}=y^{3} \), \( \sqrt[3]{x^{3}}=x \)
So the expression becomes \( \frac{5y^{3}\sqrt[3]{4y}}{x} \)

Answer:

\( \frac{5y^{3}\sqrt[3]{4y}}{x} \)