# How do you simplify (x ^ { 5} y ^ { - 9} ) ^ { 3}?

Mar 3, 2018

Since the expression is inside parenthesis, the exponent 3 will affect the entire expression

#### Explanation:

In this form, the laws of exponents tell us (a^xb^y)^n)=a^{nx}b^{ny}

So, in your case ${\left({x}^{5} {y}^{- 9}\right)}^{3} \to {x}^{5 \cdot 3} {y}^{- 9 \cdot 3} \to {x}^{15} {y}^{-} 27$

Mar 3, 2018

${x}^{15} \cdot {y}^{- 27}$
$\textcolor{red}{\left(1\right) {\left(M \cdot N\right)}^{Z} = {M}^{Z} \cdot {N}^{Z}}$
$\textcolor{red}{\left(2\right) {\left({A}^{M}\right)}^{N} = {A}^{M N}}$
${\left({x}^{5} \cdot {y}^{- 9}\right)}^{3} = {\left({x}^{5}\right)}^{3} \cdot {\left({y}^{- 9}\right)}^{3} ,$ [Applying (1)]
(x^5*y^(-9))^3=(x^((5)(3))*(y^((-9)(3))), [Applying (2)]
${\left({x}^{5} \cdot {y}^{- 9}\right)}^{3} = {x}^{15} \cdot {y}^{- 27}$