# How do you simplify (5x ^ { 3} ) ( 2x ) ^ { - 3}?

Apr 6, 2018

$\left(5 {x}^{3}\right) {\left(2 x\right)}^{- 3} = \frac{5}{8}$

#### Explanation:

When you see some term raised to a negative power, it's really shorthand for division.

In other words, ${a}^{-} n = \frac{1}{{a}^{n}}$.

So $\left(5 {x}^{3}\right) {\left(2 x\right)}^{- 3} = \frac{5 {x}^{3}}{2 x} ^ 3$.

Notice that the $2$ in the denominator is inside the parentheses, so we must raise it to the third power.

$\frac{5 {x}^{3}}{2 x} ^ 3 = \frac{5 {x}^{3}}{8 {x}^{3}}$.

At this point, we notice that ${x}^{3}$ is in the numerator and denominator, so it can be canceled. (Removed.)

$\frac{5 {x}^{3}}{8 {x}^{3}} = \frac{5}{8}$.

And $\frac{5}{8}$ cannot be further simplified, so this is our final answer.

Note: It is also appropriate to mention that $x \ne 0$, since if $x = 0$ then $\frac{5 {x}^{3}}{8 {x}^{3}} = \frac{0}{0}$ and one cannot divide by zero.

Apr 6, 2018

$\frac{5}{8}$

#### Explanation:

First, let's rewrite this without the negative exponent. The negative means that the term is in the denominator:

$\left(5 {x}^{3}\right) {\left(2 x\right)}^{-} 3 = \frac{5 {x}^{3}}{{\left(2 x\right)}^{3}}$

Now, let's distribute the power in the denominator:

$\frac{5 {x}^{3}}{{2}^{3} {x}^{3}} = \frac{5 {x}^{3}}{8 {x}^{3}}$

The last step is to cancel out the ${x}^{3}$ terms which gives us:

$\frac{5}{8}$
We can do this because any number divided by itself is 1 which means:

$\frac{5 {x}^{3}}{8 {x}^{3}} = \frac{5}{8} \cdot {x}^{3} / {x}^{3} = \frac{5}{8} \cdot 1$

We can also approach the original problem a different way.
After we distribute the $- 3$ exponent, we get the following:

$\left(5 {x}^{3}\right) {\left(2 x\right)}^{-} 3 = \left(5 {x}^{3}\right) \left({2}^{-} 3\right) \left({x}^{-} 3\right)$

Now when we simplify this equation we get:

$5 \cdot {x}^{3} \cdot \frac{1}{8} \cdot {x}^{-} 3$

After rearranging this expression we can then use the rules of exponents. When you multiply terms with the same base you add their exponents together:

$5 \cdot \frac{1}{8} \cdot {x}^{3} \cdot {x}^{-} 3 = \frac{5}{8} {x}^{3 + \left(- 3\right)} = \frac{5}{8} {x}^{0} = \frac{5}{8}$