SOLUTION: 36ab3(3 to third power)bc2(2 to 2nd power) over 24ab(4 power)c(2 power)

36ab3(3 to third power)bc2(2 to 2nd power) over 24ab(4 power)c(2 power)

I can't be sure from the above, but I think your problem might be this:

36(ab3)3(bc2)2
---------------
  24(ab)4c2

If so then first make sure every factor inside parentheses has an exponent
showing even if it is 1.

36(a1b3)3(b1c2)2
----------------
  24(a1b1)4c2

Now remove parentheses by multiplying every inner
exponent by the outer exponents:


 36a3b9b2c4
------------
  24a4b4c2

Add the exponents of the b's on top

 36a3b11c4
------------
  24a4b4c2

Divide the 36 and the 24 both by 12

 3a3b11c4
-----------
  2a4b4c2

Subtract the exponents of the a's (larger minus smaller, 4-3), placing
the result a1 in the denominator, because the denominator
contains the larger exponent 4:

 3b11c4
---------
 2a1b4c2

 3b11c4
---------
 2a1b4c2

Subtract the exponents of the b's (larger minus smaller, 11-4), placing
the result b7 in the numerator, because the numerator
contains the larger exponent 11:

 3b7c4
--------
 2a1c2

Subtract the exponents of the c's (larger minus smaller, 4-2), placing
the result c2 in the numerator, because the numerator
contains the larger exponent 4:

 3b7c2
--------
  2a1

Now you can eliminate the 1 exponent of a in the denominator

 3b7c2
-------
  2a

Edwin
AnlytcPhil@aol.com