What is #(x^-3) ^2 *x^5 / x^-1#?

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Answer 1 · 2018-03-27T16:51:55

#1#

#(x^-3) ^2 *x^5 / x^-1=x^(-3*2)*x^5*x^1=x^-6*x^6=x^6/x^6=1#

Answer 2 · 2018-03-27T16:54:50

#1#

Apply the laws of indices

#(x^-3)^2 xx x^5/x^-1" "larr (x^m)^n = x^(mn)#

#=x^-6 xx x^5/x^-1" "larr x^-m = 1/x^m#

#= 1/x^6 xx x^5 xx x^1" "larr x^m xx x^n = x^(m+n)#

#=x^6/x^6#

#=x^0" "larr x^m/x^n = x^(m-n)#

#=1" "larr x^0=1#

Answer 3 · 2018-03-27T16:57:30

#color(magenta)(=1#

#(x^(−3))^2⋅x^5/x^(−1#

#"Applying the law:"# #color(blue)((a^m)^n=a^(mn;#

#=x^(-3xx2)⋅x^5/x^(−1#

#=x^(-6)⋅x^5/x^(−1#

#"Applying the law:"# #color(blue)(a^m/a^n=a^(m-n)#

#=x^(-6)⋅x^(5-(-1))#

#=x^-6⋅x^6#

#"Applying the law:"# #color(blue)(a^m⋅a^n=a^(m+n);"#

#=x^(-6+6)#

#=x^0#

#"Applying the law:"# #color(blue)( a^0=1#

#color(magenta)(x^0=1#

#-"Hope this helps! :)"#