# How do you simplify (12x^(2)y^(-2))^(5)(4xy^(-3))^(-8)?

Oct 16, 2015

$\frac{81 {x}^{2} {y}^{14}}{64}$

Very detailed answer. The more mathematically able can skip many parts.

#### Explanation:

Look at the two parts separately then put them together afterwards
and simplify. As a side note to aid understanding: if we were to have general form of ${z}^{-} 2$ this may be written as $\frac{1}{{z}^{2}}$

Consider the first part: ${\left(12 {x}^{2} {y}^{- 2}\right)}^{5}$

Rewrite the bit inside the brackets as $\frac{12 {x}^{2}}{y} ^ 2$

So the whole part may be rewritten as:

${\left(\frac{12 {x}^{2}}{{y}^{2}}\right)}^{5}$

A method example: ${\left({x}^{2}\right)}^{3}$ is the same as ${x}^{2} \times {x}^{2} \times {x}^{2} = {x}^{2 \times 3} = {x}^{6}$

Back to the question:

${\left(\frac{12 {x}^{2}}{{y}^{2}}\right)}^{5} = \frac{{12}^{5} {x}^{2 \times 5}}{{y}^{2 \times 5}} = \frac{{12}^{5} {x}^{10}}{y} ^ 10$ ......... ( 1 )

Consider the second part: ${\left(4 x {y}^{- 3}\right)}^{- 8}$

Rewrite the bit inside the bracket as $\frac{4 x}{y} ^ \left(3\right)$

So the whole part is may be written as:

${\left(4 \frac{x}{y} ^ \left(3\right)\right)}^{- 8}$

Although it is not normally written like this it gives:
$\frac{1}{\frac{4 x}{{y}^{3}}} ^ 8$

The net result of this is:

${\left(\frac{{y}^{3}}{4 x}\right)}^{8} = \frac{{y}^{24}}{{4}^{8} {x}^{8}}$ ....... ( 2 )

Putting (1) and (2) together gives:

$\frac{{12}^{5} {x}^{10}}{y} ^ 10 \times \frac{{y}^{24}}{{4}^{8} {x}^{8}}$

Grouping variables and constants we have:

( remember that ${12}^{5} = {\left(3 \times 4\right)}^{5} = {3}^{5} \times {4}^{5}$)

so we have:

$\frac{{3}^{4} \times {4}^{5}}{4} ^ 8 \times {x}^{10} / {x}^{8} \times \frac{{y}^{24}}{{y}^{10}}$

Simplifying -> $\frac{81}{64} \times {x}^{2} \times {y}^{14}$

$\frac{81 {x}^{2} {y}^{14}}{64}$