Question

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

Answer 1

`x^2/y^-3(x^2/y-(3z^-2y)/x)=x^4-(xy^4)/(9z^2)`

Explanation:

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

Let's start with the original problem:

`x^2/y^-3(x^2/y-(3z^-2y)/x)`

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

`(x^2/y^-3)(x^2/y)-(x^2/y^-3)(3z^-2y/x)`

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*a^c=a^(b+c)`

2) `a^b/c^b=(a-c)^b`

3) `a^-b=1/a^b`

`(x^2*x^2)/(y^-3*y^1)-(x^2/y^-3)(3z^-2y/x)`

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

`x^4/y^-2=x^4/(1/y^2)=x^4*y^2=x^4y^2`

Now let's work on the right side:

`(x^2/y^-3)(3z^-2y/x)=(x^2/(1/y^3))(1/((3z)^2)*y/x)=(x^2y^3)(y/(9xz^2))=(xy^3)(y/(9z^2))=(xy^4)/(9z^2)`

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

`x^4-(xy^4)/(9z^2)`

And, you're done!