find the product or quotient and simply?

Question

4x-8ax^2\3 divided by 8a^2x^2-2 \3+6ax..could you show the work..TY..I don't have an example in the textbook I'm using

problem should have read as follows:
4x-8ax^2 over 3 divided by 8a^2-2 over 3+6ax

Answer 1

First, turn the second expression 8a...etc. upside down and change the division sign to multipy. Now, you can expand the top and bottom turns. Some people call it the "inside out rule" Anyway, you multiply each of the terms in the first (4x and -8ax^2) by each of the terms in the second expression (3 + 6ax)... First term times first term is 4x*3 = 12x. First term times second term is 4x*6ax = 24ax^2. and so on. Now I have
(12x + 24ax^2 - 24ax^2 - 48a^2x^3)/(24a^2x^2-6). Now I notice that all terms are divisible by 6. So I just divide everything through by 6. I also notice that all the terms on top have an "x" so I will factor out x. Now I have
x*(2-4ax-4ax-8a^2x^2)/(4a^2x^2-1)
I can add the two 4ax terms in the top together. When I see that bottom with a first term that is a square and a second term that is a negative one, I know that I can factor the bottom. This gives
(x*(2-8ax-8a^2x^2)/((2ax+1)*(2ax-1))
Now I can make a guess that maybe (2ax+1) or (2ax-1) is a factor of the top part. After fiddling around, you can factor the top part into -2x*(2ax+1)*(2ax+1). Now I can cross off (2ax+1) from the top and bottom. This leaves me with
-2x*(2ax+1)/(2ax-1)

Answer 2

do u mean the numbers as

(4x -8ax⅔) / (8a² x² -⅔ +6ax)

There appears to be some errors. Pl check.