Hello good people!!!..Juriemagic here. First of all to all of you guys out there, sitting ready to help those in need, I wish each and every one a very happy

${(p^2q)^3(2p^3q^-4)^2}\over{p^2q^3*(4p^-2q^5)^3}$

I'm ready to help with problems written in English,JK! I'm sorry, I can't help u😥

Simplify the following:
((p^2 q)^3 ((2 p^3)/q^4)^2)/(p^2 q^3 ((4 q^5)/p^2)^3)

Multiply each exponent in p^2 q by 3:
(p^(3×2) q^3 ((2 p^3)/q^4)^2)/(p^2 q^3 ((4 q^5)/p^2)^3)

3×2  =  6:
(p^6 q^3 ((2 p^3)/q^4)^2)/(p^2 q^3 ((4 q^5)/p^2)^3)

Multiply each exponent in (2 p^3)/q^4 by 2:
(p^6 q^3×(2^2 p^(2×3))/((q^4)^2))/(p^2 q^3 ((4 q^5)/p^2)^3)

2×3  =  6:
(2^2 p^6 q^3 p^6)/((q^4)^2 p^2 q^3 ((4 q^5)/p^2)^3)

2^2 = 4:
(4 p^6 q^3 p^6)/((q^4)^2 p^2 q^3 ((4 q^5)/p^2)^3)

Multiply exponents. (q^4)^2 = q^(4×2):
(4 p^6 q^3 p^6)/(q^(4×2) p^2 q^3 ((4 q^5)/p^2)^3)

4×2  =  8:
(4 p^6 q^3 p^6)/(q^8 p^2 q^3 ((4 q^5)/p^2)^3)

Multiply each exponent in (4 q^5)/p^2 by 3:
(4 p^6 q^3 p^6)/(q^8 p^2 q^3×(4^3 q^(3×5))/((p^2)^3))

3×5  =  15:
(4 p^6 q^3 p^6)/(q^8×(4^3 p^2 q^3 q^15)/(p^2)^3)

Multiply exponents. (p^2)^3 = p^(2×3):
(4 p^6 q^3 p^6)/(q^8×(4^3 p^2 q^3 q^15)/p^(2×3))

2×3  =  6:
(4 p^6 q^3 p^6)/(q^8×(4^3 p^2 q^3 q^15)/p^6)

4^3 = 4×4^2:
(4 p^6 q^3 p^6)/(q^8×(4×4^2 p^2 q^3 q^15)/p^6)

4^2 = 16:
(4 p^6 q^3 p^6)/(q^8×(4×16 p^2 q^3 q^15)/p^6)

4×16  =  64:
(4 p^6 q^3 p^6)/(q^8×(64 p^2 q^3 q^15)/p^6)

(p^6 q^3×4 p^6)/(q^8 (p^2 q^3×64 q^15)/p^6) = q^3/q^3×((p^6×4 p^6)/q^8)/((p^2×64 q^15)/p^6) = ((p^6×4 p^6)/q^8)/((p^2×64 q^15)/p^6):
((4 p^6 p^6)/q^8)/((64 p^2 q^15)/p^6)

Combine powers. (p^6×4 p^6)/(q^8 (p^2×64 q^15)/p^6) = (p^(6+6-2) 4 p^6)/(q^8×64 q^15):
(4 p^6+6-2 p^6)/(64 q^8 q^15)

6+6-2 = 10:
(4 p^10 p^6)/(64 q^8 q^15)


4 | 1 | 6
| 6 | 4
- | 4 |
| 2 | 4
- | 2 | 4
|  | 0:
(p^10 p^6)/(16 q^8 q^15)

Combine powers. (p^10 p^6)/(16 q^8 q^15) = (p^(10+6) q^(-8-15))/16:
(p^10+6 q^-8-15)/16

10+6 = 16:
(p^16 q^(-8-15))/16

-8-15 = -23:
Answer: | (p^16 q^-23)/16

((p^2q)^3(2p^3q^-4)^2) / p^2q^3 * (4p^-2q^5)^3

This is your question - is it what you intended or do you need more brackets?

$\frac{(p^2q)^3(2p^3q^{-4})^2 }{p^2}\times q^3 * (4p^{-2}q^5)^3$

The intention is everything to the left of the division sign is on top, and everything right from it, at the bottom...

Hi Juriemagic,

I see you still cannot log on    

$\frac {(p^2q)^3(2p^3q^{-4})^2 } {p^2q^3 * (4p^{-2}q^5)^3}\\~\\ =\frac {p^6q^3(2^2p^6q^{-8}) } {p^2q^3 * (4^3p^{-6}q^{15})}\\~\\ =\frac {p^6(4p^6q^{-8}) } {p^2 * (64p^{-6}q^{15})}\\~\\ =\frac {p^4(4p^6q^{-8}) } { (64p^{-6}q^{15})}\\~\\ =\frac {4p^4p^6q^{-8} } { 64p^{-6}q^{15}}\\~\\ =\frac {p^{10}q^{-8} } { 16p^{-6}q^{15}}\\~\\ =\frac {p^{10}p^{+6} } { 16q^{15}q^{+8}} \\~\\ =\frac {p^{16} } { 16q^{23}} \\ $

If you do not understand or you want a particular line discussed then please ask :)

Hi Melody!,

Thank you sooo much. I managed to do this sum but only to a point...

Yes, the logging on is still a problem, aw well, I'll speak with I.T. here and see if they can help...otherwise?..I trust you are all okay?

Hi Juriemagic,

Admin knows that you and others have log in problems.  He does know how to fix it.

Have you tried using a different browser?  

Do you completely understand my answer?

You should thank guest answerer too.  He/she put in a lot of work on your behalf.   :/

Here are some more examples to practic on if you want

http://web2.0calc.com/questions/indices-especially-negative-indices