# How do you simplify (20x^5)/y^2 * ((x^2y^2)/(2x))^3?

Oct 26, 2015

$\frac{5}{2} {x}^{8} {y}^{4}$

#### Explanation:

First of all, expand the power of the second factor: since the power of a fraction is the power of the numerator divided by the power of the denominator, we have that

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

Now, the power of a product is again the power of every single factor, so

${\left({x}^{2} {y}^{2}\right)}^{3} = {\left({x}^{2}\right)}^{3} \cdot {\left({y}^{2}\right)}^{3}$, and ${\left(2 x\right)}^{3} = {2}^{3} \cdot {x}^{3}$.

The last rule we need is the one which states that when we deal with the power of a power, we must multiplicate the exponents:

${\left({x}^{2}\right)}^{3} = {x}^{2 \cdot 3} = {x}^{6}$, and the same goes for ${\left({y}^{2}\right)}^{3}$

The result is thus

${\left(\frac{{x}^{2} {y}^{2}}{2 x}\right)}^{3} = \frac{{x}^{6} {y}^{6}}{8 {x}^{3}}$.

Now we're ready to multiply and cross simplify:

$\frac{\textcolor{g r e e n}{20} {x}^{5}}{\textcolor{red}{{y}^{2}}} \cdot \frac{\textcolor{b l u e}{{x}^{6}} \textcolor{red}{{y}^{6}}}{\textcolor{g r e e n}{8} \textcolor{b l u e}{{x}^{3}}}$

So, we can simplify $20$ and $8$ dividing both by $4$, obtaining $5$ and $2$.

Also, ${y}^{2}$ cancels out and ${y}^{6}$ becomes ${y}^{4}$, and ${x}^{3}$ cancels out too and ${x}^{6}$ becomes ${x}^{3}$.

So, we're left with

$5 {x}^{5} \cdot \frac{{x}^{3} {y}^{4}}{2} = \frac{5 {x}^{8} {y}^{4}}{2}$