How do you write #((-3n^2)/5)^-3# in simplest form?

1 Answer
Feb 23, 2017

First, use this rule for exponents: #a = a^color(red)(1)#

#((-3n^2)/5)^-3 = ((-3^color(red)(1)n^color(red)(2))/5^color(red)(1))^-3#

Next, use this rule of exponents: #(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))#

#((-3^color(red)(1)n^color(red)(2))/5^color(red)(1))^color(blue)(-3) = (-3^(color(red)(1)xxcolor(blue)(-3))n^(color(red)(2)xxcolor(blue)(-3)))/5^(color(red)(1)xxcolor(blue)(-3)) = (-3^-3n^-6)/5^-3#

Now, we can use these rule of exponents to eliminate the negative exponents:

#x^color(red)(a) = 1/x^color(red)(-a)# and #1/x^color(red)(a) = x^color(red)(-a)#

#(-3^color(red)(-3)n^color(red)(-6))/5^color(red)(-3) = 5^color(red)(- -3)/(-3^color(red)(- -3)n^color(red)(- -6)) = 5^color(red)(3)/(-3^color(red)(3)n^color(red)(6)) = -125/(27n^6)#