# How do you simplify  (x^-4y^-6z^-10 )/( a^1b^2c^-4)^2 * (a^1b c^-4) /( x^6y z^9)^2?

Sep 20, 2015

$\frac{{c}^{4}}{{x}^{16} \cdot {y}^{8} \cdot {z}^{28} \cdot {a}^{1} \cdot {b}^{3}}$

#### Explanation:

As always, "count apples as apples and oranges as oranges".
In this case everything are really just multiplications and divisions.
If I see this, I would just open up the parentheses and see what can be removed. The remaining terms have to be the answer.

So let's attack this.
We have:
$\frac{{x}^{-} 4 {y}^{-} 6 {z}^{-} 10}{{\left({a}^{1} {b}^{2} {c}^{-} 4\right)}^{2}} \cdot \frac{{a}^{1} b {c}^{-} 4}{{\left({x}^{6} y {z}^{9}\right)}^{2}}$

Then, I would rewrite $b = {b}^{1}$ (although it is unnecessary), and $y = {y}^{1}$ just to be explicit.
Open the parentheses. Remember the rule: ${\left({a}^{n}\right)}^{m} = {a}^{n \cdot m}$
We get:
$\frac{{x}^{-} 4 {y}^{-} 6 {z}^{-} 10}{{a}^{2} {b}^{4} {c}^{-} 8} \cdot \frac{{a}^{1} {b}^{1} {c}^{-} 4}{{x}^{12} {y}^{2} {z}^{18}}$

Then it's easy, x goes with x, y goes with y, etc...
Remember the rule $\frac{{a}^{n}}{{a}^{m}} = {a}^{n - m}$
So we have:
${x}^{- 4 - 12} \cdot {y}^{- 6 - 2} \cdot {z}^{- 10 - 18} \cdot {a}^{1 - 2} \cdot {b}^{1 - 4} \cdot {c}^{- 4 + 8}$
$= {x}^{- 16} \cdot {y}^{- 8} \cdot {z}^{- 28} \cdot {a}^{- 1} \cdot {b}^{- 3} \cdot {c}^{4}$
or rewriting it in fractional form (putting all the negative exponents at the bottom):

$= \frac{{c}^{4}}{{x}^{16} \cdot {y}^{8} \cdot {z}^{28} \cdot {a}^{1} \cdot {b}^{3}}$