What is #((2x^0*2x^3)/(xy^-4))^-3#?

2 Answers
Jul 28, 2016

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

Explanation:

#((2x^0. 2x^3)/(xy^-4))^-3#
Since #x^0=1# we get
#((2(1). 2x^3)/(xy^-4))^-3#
#=((4x^3)/(xy^-4))^-3#
#=((4x^2)/(y^-4))^-3#
#=((4x^2)(y^4))^-3#
#=(4x^2y^4)^-3#
#=1/(4x^2y^4)^3#

Jul 28, 2016

#1/(64x^6y^12)#

Explanation:

There are a number of laws of indices going on here.

No law is more important than another. There are different ways of simplifying the expression.

#((2x^0xx 2x^3)/(xy^-4))^-3 " Look for the obvious laws first"#

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

=#((2xxcolor(red)(1)xx2color(blue)(x^2))/y^-4)^(-3)#

=#(color(green)(2xx2x^2)/color(orange)(y^-4))^color(magenta)(-3)" "(a/b)^-m = (b/a)^(+m)#

=#(color(orange)(y^-4)/color(green)(2xx2x^2))^color(magenta)3#

=#(1/(2xx2x^2color(orange)(y^4)))^3" "color(orange)(x^-1 = 1/x)#

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

=#color(red)(1/(64x^6y^12))#