How do you simplify #(\frac { 2x ^ { 0} y ^ { 3} } { x ^ { - 1} y ^ { 4} } ) ^ { 3}#?

1 Answer
Nov 13, 2017

See a solution process below:

Explanation:

First, we can use this rule of exponents to simplify the #x# terms within the parenthesis:

#x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))#

#((2x^color(red)(0)y^3)/(x^color(blue)(-1)y^4))^3 =>#

#((2x^(color(red)(0)-color(blue)(-1))y^3)/y^4)^3 =>#

#((2x^(color(red)(0)+color(blue)(1))y^3)/y^4)^3 =>#

#((2x^1y^3)/y^4)^3#

Next, use this rule of exponents to simplify the #y# term within the parenthesis:

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

#((2x^1y^color(red)(3))/y^color(blue)(4))^3 =>#

#((2x^1)/y^(color(blue)(4)-color(red)(3)))^3 =>#

#((2x^1)/y^1)^3#

Use this rule of exponents to rewrite the constant:

#a = a^color(red)(1)#

#((2^1x^1)/y^1)^3#

Now, use this rule of exponents to complete the simplification:

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

#((2^color(red)(1)x^color(red)(1))/y^color(red)(1))^color(blue)(3) =>#

#(2^(color(red)(1)xxcolor(blue)(3))x^(color(red)(1)xxcolor(blue)(3)))/y^(color(red)(1)xxcolor(blue)(3)) =>#

#(2^3x^3)/y^3 =>#

#(8x^3)/y^3#