Answer:
[tex]\dfrac{m^3}{16p^2q^2}[/tex]
Step-by-step explanation:
Given:
[tex](2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}[/tex]
1.
[tex]m^{-1}=\dfrac{1}{m}[/tex]
2.
[tex]q^0=1[/tex]
3.
[tex]2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}[/tex]
4.
[tex](2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}[/tex]
5.
[tex]m^{-1}=\dfrac{1}{m}[/tex]
6.
[tex]2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}[/tex]
7.
[tex]\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}[/tex]
8.
[tex](2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}[/tex]
