How do you simplify #((5x)/(3yz))^-3# and write it using only positive exponents?

1 Answer
Jan 14, 2018

Answer:

See a solution process below:

Explanation:

First, let's use this rule of exponents to rewrite the expression as:

#1/4y + (1/4 xx 3/2) =>#

#((5x)/(3yz))^color(red)(-3) => 1/((5x)/(3yz))^color(red)(- -3) => 1/((5x)/(3yz))^3#

Now, use these rules of exponents to eliminate the outer exponent:

#a = a^color(red)(1)# and #(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))#

#1/((5^color(red)(1)x^color(red)(1))/(3^color(red)(1)y^color(red)(1)z^color(red)(1)))^color(blue)(3) => 1/((5^(color(red)(1)xxcolor(blue)(3))x^(color(red)(1)xxcolor(blue)(3)))/(3^(color(red)(1)xxcolor(blue)(3))y^(color(red)(1)xxcolor(blue)(3))z^(color(red)(1)xxcolor(blue)(3)))) => 1/((5^3x^3)/(3^3y^3z^3)) => 1/((125x^3)/(27y^3z^3)) => (27y^3z^3)/(125x^3)#