Question

(2x^3 divide 3y^4)^2 multiply (x divide 3y)^-2

Answer 1

To simplify the expression \((\frac{2x^3}{3y^4})^2 \times \left(\frac{x}{3y}\right)^{-2}\), follow these steps:

1. **Square the first part**:
\[
\left(\frac{2x^3}{3y^4}\right)^2 = \frac{(2x^3)^2}{(3y^4)^2} = \frac{4x^6}{9y^8}
\]

2. **Apply the negative exponent on the second part**:
\[
\left(\frac{x}{3y}\right)^{-2} = \frac{1}{\left(\frac{x}{3y}\right)^2} = \frac{1}{\frac{x^2}{9y^2}} = \frac{9y^2}{x^2}
\]

3. **Multiply the results from steps 1 and 2**:
\[
\frac{4x^6}{9y^8} \times \frac{9y^2}{x^2} = \frac{4x^6 \cdot 9y^2}{9y^8 \cdot x^2}
\]

The \(9\) in the numerator and denominator cancels out:
\[
= \frac{4x^6y^2}{y^8x^2}
\]

4. **Combine the terms**:
\[
= \frac{4x^{6-2}}{y^{8-2}} = \frac{4x^4}{y^6}
\]

Thus, the simplified expression is:

\[
\boxed{\frac{4x^4}{y^6}}
\]