The problem asks us to simplify the expression $(\frac{2x^3}{y})^{-3}$ and express the answer with positive exponents.

1. Problem Description

The problem asks us to simplify the expression $(\frac{2x^3}{y})^{-3}$ and express the answer with positive exponents.

2. Solution Steps

We are given the expression $(\frac{2x^3}{y})^{-3}$.

We can apply the power of a quotient rule, which states that $(\frac{a}{b})^n = \frac{a^n}{b^n}$. Thus, we have:

$(\frac{2x^3}{y})^{-3} = \frac{(2x^3)^{-3}}{y^{-3}}$

Next, we apply the power of a product rule, which states that $(ab)^n = a^n b^n$. Therefore, $(2x^3)^{-3} = 2^{-3} (x^3)^{-3}$.

$\frac{(2x^3)^{-3}}{y^{-3}} = \frac{2^{-3}(x^3)^{-3}}{y^{-3}}$

Now, we apply the power of a power rule, which states that $(a^m)^n = a^{mn}$. Therefore, $(x^3)^{-3} = x^{3 \cdot (-3)} = x^{-9}$.

$\frac{2^{-3}x^{-9}}{y^{-3}}$

We want to express the answer with positive exponents. Recall that $a^{-n} = \frac{1}{a^n}$. Therefore, $2^{-3} = \frac{1}{2^3} = \frac{1}{8}$, $x^{-9} = \frac{1}{x^9}$, and $y^{-3} = \frac{1}{y^3}$.

$\frac{2^{-3}x^{-9}}{y^{-3}} = \frac{\frac{1}{2^3} \cdot \frac{1}{x^9}}{\frac{1}{y^3}} = \frac{\frac{1}{8x^9}}{\frac{1}{y^3}} = \frac{1}{8x^9} \cdot \frac{y^3}{1} = \frac{y^3}{8x^9}$.

Final Answer:

3. Final Answer

$\frac{y^3}{8x^9}$