# How do you simplify \frac { x ^ { 2} } { y ^ { - 3} } ( \frac { x ^ { 2} } { y } - \frac { 3z ^ { - 2} y } { x } )?

Oct 21, 2017

${x}^{2} / {y}^{-} 3 \left({x}^{2} / y - \frac{3 {z}^{-} 2 y}{x}\right) = {x}^{4} - \frac{x {y}^{4}}{9 {z}^{2}}$

#### Explanation:

This may seem like a complicated problem due to all the variables and exponents. But, it's not.

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

Let's first use the Distributive Property to simplify things a little bit:

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

We can then use some rules about multiplying together exponents with the same base as well as negative exponents to simplify a little bit more:

1) ${a}^{b} \cdot {a}^{c} = {a}^{b + c}$

2) ${a}^{b} / {c}^{b} = {\left(a - c\right)}^{b}$

3) ${a}^{-} b = \frac{1}{a} ^ b$

$\frac{{x}^{2} \cdot {x}^{2}}{{y}^{-} 3 \cdot {y}^{1}} - \left({x}^{2} / {y}^{-} 3\right) \left(3 {z}^{-} 2 \frac{y}{x}\right)$

Let's take care of the left side of the expression first:

${x}^{4} / {y}^{-} 2 = {x}^{4} / \left(\frac{1}{y} ^ 2\right) = {x}^{4} \cdot {y}^{2} = {x}^{4} {y}^{2}$

Now let's work on the right side:

$\left({x}^{2} / {y}^{-} 3\right) \left(3 {z}^{-} 2 \frac{y}{x}\right) = \left({x}^{2} / \left(\frac{1}{y} ^ 3\right)\right) \left(\frac{1}{{\left(3 z\right)}^{2}} \cdot \frac{y}{x}\right) = \left({x}^{2} {y}^{3}\right) \left(\frac{y}{9 x {z}^{2}}\right) = \left(x {y}^{3}\right) \left(\frac{y}{9 {z}^{2}}\right) = \frac{x {y}^{4}}{9 {z}^{2}}$

After all that, let's put them together, while not forgetting the $-$ sign in the middle:

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

And, you're done!