Question 597185
You are right in thinking that you need to do some factoring and taking out a 4 from each denominator is a good start.
I'll show you the next steps, and what clues and tricks I used.
 
THE SOLUTION (skipping explanations for now):
{{{(x+3y)/(4x^2+12xy+8y^2)-(x+y)/(4x^2+20xy+24y^2)+(x+2y)/( 4x^2+16xy+12y^2)}}}
={{{(x+3y)/(4(x^2+3xy+2y^2))-(x+y)/(4(x^2+5xy+6y^2))+(x+2y)/(4(x^2+4xy+3y^2))}}}
 
={{{(x+3y)/(4(x+y)(x+2y))-(x+y)/(4(x+2y)(x+3y))+(x+2y)/(4(x+y)(x+3y))}}}
 
={{{((x+3y)^2-(x+y)^2+(x+2y)^2)/(4(x+y)(x+2y)(x+3y))}}}
 
={{{((x^2+6xy+9y^2) -(x^2+2xy+y^2)+(x^2+4xy+4y^2))/(4(x+y)(x+2y)(x+3y))}}}
 
={{{(x^2+6xy+9y^2-x^2-2xy-y^2+x^2+4xy+4y^2)/(4(x+y)(x+2y)(x+3y))}}}
 
={{{(x^2+8xy+12y^2)/(4(x+y)(x+2y)(x+3y))}}}
 
={{{((x+2y)(x+6y))/(4(x+y)(x+2y)(x+3y))}}}
 
={{{(x+6y)/(4(x+y)(x+3y))}}}
 
ABOUT FACTORING:
When you have a polynomial with x and y, like
{{{x^2+3xy+2y^2}}}
factoring may look complicated, but there are clues and tricks that helped me factor it.
I noticed that he total degree (in x and y) of all the terms is 2, so I knew that
{{{x^2+3xy+2y^2=y^2(x^2/y^2+3x/y+2)=y^2((x/y)^2+3(x/y)+2)}}}
and renaming variables using {{{z=x/y}}} the big bracket turns into {{{z^2+3z+2}}}
I knew I could handle that.
It is easier to factor when there is only one variable, and if it seemed difficult I could always resort to the quadratic formula to find that the zeros of that polynomial. It would tell me that the zeros are z=-1 and z=-2, and from there I would figure out that the factors are (z+1) and (z+2).