#color(blue)("Preamble")#
Consider the context of #n^0#
#" "#suppose we had #2^3/2^3 =1#
#" "#This can be written as #2^(3-3) = 2^0=1#
#" "# so #n^0# disappears in the multiplication
....................................................................................
Consider the context of #((m^3p^5)/n^7)^6#
Suppose we had #p^2# then this is #pxxp#
So whatever #p# is it is multiplied by itself twice.
Set #p=3^2# then we have #3^2xx3^2 = 3xx3xx3xx3 = 3^4#
So #(3^2)^2=3^(2xx2)=3^4#
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#color(blue)("Answering the question - applying the above principles")#
Given:#" "((m^3p^5)/n^7)^6xx((m^2n^0p^3)/(m^4n^2))^3#
#=(m^(18)p^(30))/n^42 xx (m^6p^9)/(m^(12)n^6)#
#=(m^(18+6)color(white)(.)p^(30+9))/(m^12color(white)(.)n^(42+6))" "# this is the same as: #" "(m^12xxcancel(m^12))/(cancel(m^12)) xx(p^39)/(n^(48)#
#=(m^12color(white)(.)p^(39))/n^(48)#
