# How do you simplify and write ((3x^-2 y^3) /(2xy))^-2 with positive exponents?

Jun 3, 2016

$\frac{4 {x}^{6}}{9 {y}^{4}}$

#### Explanation:

We start with ${\left(\frac{3 {x}^{-} 2 {y}^{3}}{2 x y}\right)}^{-} 2$. The first thing I like to do is to simplify everything that I can, which means getting rid of that external $- 2$. To do that, we just need to distribute it in. That means we now have ${\left(3 {x}^{-} 2 {y}^{3}\right)}^{-} \frac{2}{2 x y} ^ - 2$ or $\frac{\frac{1}{9} {x}^{4} {y}^{-} 6}{\frac{1}{4} {x}^{-} 2 {y}^{-} 2}$.

First thing's first, let's get rewrite $\frac{\frac{1}{9}}{\frac{1}{4}}$. To solve a multiplication problem with fractions, we just need to mutiply the recipricol. For us, that means $\frac{1}{9} \cdot \frac{4}{1}$. Multiply straight across to give us $\frac{4}{9}$.

Now, in order to change the sign of the power, we need to move it into the numerator denominator. What I mean is, if we have a negative power in the denominator, like $\frac{1}{x} ^ - 2$, we need to move the component to to the numerator to change the sign. If we do that, we get ${x}^{2} / 1$, which is now positive.

For this problem, that means moving the components around, like this: $\frac{4}{9} \left({x}^{4} \cdot {x}^{2} \cdot {y}^{2} \cdot {y}^{-} 6\right)$. Now, you may notice that I haven't moved ${y}^{-} 6$. that's because I want to combine like terms before I do that. So, ${x}^{4} \cdot {x}^{2}$ becomes ${x}^{6}$ because exponents add when the bases are multiplied. That also means that ${y}^{2} \cdot {y}^{-} 6$ is ${y}^{4}$.

We should now have $\frac{4 {x}^{6} {y}^{-} 4}{9}$. Let's chnage the sign on that ${y}^{-} 4$ to get $\frac{4 {x}^{6}}{9 {y}^{4}}$. That's it! We are done.

Jun 3, 2016

$\frac{4 {x}^{6}}{9 {y}^{4}}$

#### Explanation:

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

There is a nifty rule when working with indices which states that if a fraction is raised to a negative index, if you find the reciprocal of the fraction the index becomes positive:

${\left(\frac{x}{y}\right)}^{-} m \text{ becomes } {\left(\frac{y}{x}\right)}^{+ m}$

Therefore in the question posed, we can get the following:

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

Now square everything in the bracket:

$\frac{4 {x}^{2} {y}^{2}}{9 {x}^{-} 4 {y}^{6}}$

Simplifying the indices gives the final answer:

$\frac{4 {x}^{6}}{9 {y}^{4}} \text{ note that 2 -(-4) = 6}$