Reduce (64y^3+27z^3)/(4y+3y) 

 (4y)² + (3z)²
---------------
    4y + 3z 

to lowest terms choices are

16y² + 12yz - 9z²

4y + 3z

16y² - 12yz + 9z²

already in lowest terms


-----------------------------------------
Notice that I have placed parentheses around
both the numerator and the denominator as must
be done when typing an algebraic fraction all
on one line whenever there is more than just 
one number or only one letter in the numerator 
or denominator.  You of course don't need
parentheses when typing it on two separate
lines like this:


 64y³ + 27z³
-------------
   4y + 3y 

The numerator can be rewritten as

 (4y)² + (3z)²
---------------
    4y + 3z 

Factor the numerator, (4y)³ + (3z)³,
using the principle:

A³ + B³ = (A + B)(A² - AB + B²)

where A = 4y and B = 3z, so we have

Let's isolate the numerator to work with it.

(4y)² + (3z)² = 

(4y + 3z)[(4y)² - (4y)(3z) + (3z)²]

=  (4y + 3z)(16y² - 12yz + 9z²)

Now place it over the denominator

  (4y + 3z)(16y² - 12yz + 9z²)
--------------------------------
            4y + 3z

Now we cancel the entire denominator into the
first factor of the numerator:

      1
  (4y + 3z)(16y² - 12yz + 9z²)
--------------------------------
            4y + 3z
               1

All that's left is

        16y² - 12yz + 9z²
  
which is the third choice.

Edwin