How do you simplify [(8(x^3) ^2 y^-3)/(6y^2 x^-2)] [(4(x^2 ) ^2 y^-3)/(12x^-3 y^2 )] using only positive exponents?

Apr 21, 2016

$\frac{4 {x}^{15}}{9 {y}^{10}}$

Explanation:

Using the following $\textcolor{b l u e}{\text{ rules of exponents }}$

• a^mxxa^n = a^(m+n)

example : ${2}^{2} \times {2}^{3} = {2}^{2 + 3} = {2}^{5} = 32$
$\textcolor{red}{\text{-------------------------------------------------}}$
•( a^m)/(a^n) = a^(m-n)
example$\frac{{3}^{4}}{{3}^{2}} = {3}^{4 - 2} = {3}^{2} = 9$
and $\frac{{2}^{3}}{{2}^{5}} = {2}^{3 - 5} = {2}^{- 2}$
$\textcolor{red}{\text{--------------------------------------------------}}$
• a^(-m) hArr 1/(a^m)
example (from above): $\frac{{2}^{3}}{{2}^{5}} = {2}^{- 2} = \frac{1}{{2}^{2}} = \frac{1}{4}$
$\textcolor{red}{\text{----------------------------------------------------}}$
• (a^m)^n = a^(mxxn)= a^(mn)

example${\left({2}^{2}\right)}^{3} = {2}^{2 \times 3} = {2}^{6} = 64$
$\textcolor{red}{\text{-----------------------------------------------------}}$

now applying these to the question.

first bracket$\left[\frac{8 {\left({x}^{3}\right)}^{2} {y}^{- 3}}{6 {y}^{2} {x}^{-} 2}\right] = \frac{8}{6} \times \frac{{x}^{6}}{{x}^{-} 2} \times \frac{{y}^{-} 3}{y} ^ 2$

$= \frac{4}{3} {x}^{6 - \left(- 2\right)} {y}^{- 3 - 2} = \frac{4}{3} {x}^{8} {y}^{-} 5$
$\textcolor{b l u e}{\text{-------------------------------------------------}}$

second bracket $\left[\frac{4 {\left({x}^{2}\right)}^{2} {y}^{-} 3}{12 {x}^{-} 3 {y}^{2}}\right] = \frac{4}{12} \times {x}^{4} / {x}^{-} 3 \times {y}^{-} \frac{3}{y} ^ 2$

$= \frac{1}{3} {x}^{7} {y}^{-} 5$
$\textcolor{b l u e}{\text{------------------------------------------------}}$

Now multiplying the 2 simplifications together

 4/3 x^8y^-5xx1/3 x^7y^-5 =4/9x^(15)y^(-10)=(4x^(15))/(9y^(10)