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
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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)