Maths C3 - Simplifying Algebraic Fractions...Help with tricky Question?

Ok so I've just spent a long time trying to work out Exercise 1C question m from the Edexcel C3 Modular textbook. I managed to solve the answer but took so many steps to get it! There must be a quicker way to solve this question. The question is...

Question m..:
$\frac{3}{x^2+3x+2}-\frac{2}{x^2+4x+4}$

The steps I took to work out the answer are as follows...
$\frac{3}{(x+1)(x+2)}-\frac{2}{(x+2)(x+2)}$
...
$\frac{3x^2+12x+12}{(x+1)(x+2)^3}-\frac{2x^2+6x+4}{(x+1)(x+2)^3}$
...
$\frac{x^2+6x+8}{(x+1)(x+2)^3}$
...
$\frac{(x+2)(x+4)}{(x+1)(x+2)^3}$
...
$\frac{x+4}{(x+1)(x+2)^2}$

Use a common denominator of $(x+1)(x+2)^2$ instead of
$(x+1)(x+2)^3$

How did you figure out the common denominator so easily?
Please explain as I'm terrible at working out common denominators

How did you figure out the common denominator so easily?
Please explain as I'm terrible at working out common denominators

How did you figure out the common denominator so easily?
Please explain as I'm terrible at working out common denominators

How did you figure out the common denominator so easily?
Please explain as I'm terrible at working out common denominators

Seems you have a gap somewhere in your knowledge...

Watch these videos:
https://www.youtube.com/watch?v=moLdgkGjrWI
https://www.youtube.com/watch?v=mRSgfo6hZlo

Never be afraid to revisit the basics...

You don't need $(x+2)^3$ as (x+2) is a factor of $(x+2)^2$ it's the same with numbers, you're looking for the lowest common multiple.

You just do it by inspection. Look what you need in order to make the denominators the same and that's it. Practice some more questions and you'll start to recognise them quicker.