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Assuming that\(p \neq 0\)  and\(q \neq 0\) , simplify \(\dfrac{(pq^3)^3(4p^2q)^2}{(2pq^2)^3}\)    plz include a step by step explanation

 Jun 19, 2020
 #1
avatar+8342 
0

You should know that 

\(p^a\cdot p^b = p^{a + b} \qquad (1)\\ \dfrac{p^a}{p^b} = p^{a - b}\qquad (2)\\ \left(p^a\right)^b = p^{ab}\qquad (3)\\ (pq)^a = p^a q^a\qquad (4)\)

 

Using these three formulae,

\(\quad \dfrac{\left(pq^3\right)^3\cdot \left(4p^2q\right)^2}{\left(2pq^2\right)^3}\\ \stackrel{(4)}{=} \dfrac{p^3 \left(q^3\right)^3\cdot 4^2 \cdot \left(p^2\right)^2\cdot q^2}{2^3 \cdot p^3 \cdot \left(q^2\right)^3}\\ \stackrel{(3)}{=} \dfrac{p^3q^9\cdot16\cdot p^4\cdot q^2}{8p^3q^6}\\ = 2\cdot \dfrac{p^3\cdot p^4}{p^3} \cdot \dfrac{q^9 \cdot q^2}{q^6}\)

 

Can you do the rest with identities (1) and (2)?

 Jun 19, 2020
 #2
avatar+130 
-4

so would it be \(2p^{4}\) times \(2q^{11}\)

Creampuff  Jun 19, 2020
 #3
avatar+8342 
0

No. The 2 shouldn't be distributed to each fraction, and you did the "q" one wrong.

MaxWong  Jun 19, 2020

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