Can someone check my work to see if it's right please?

First I found the least common denominator which is x(x+1)(x+2).
I multiplied the first fraction by (x+1)(x+2)/(x+1)(x+2).
I multiplied the second fraction by x(x+2).
I multiplied the last fraction by x(x+1).

1st fraction's numerator:3x^2+9x+6.
2nd fractions numerator: 3x^2+6x.
3rd fractions numerator:3x^2+3x.

Question

Can someone check my work to see if it's right please?

  First I found the least common denominator which is x(x+1)(x+2).
I multiplied the first fraction by (x+1)(x+2)/(x+1)(x+2).
I multiplied the second fraction by x(x+2).
I multiplied the last fraction by x(x+1).

1st fraction's numerator:3x^2+9x+6.
2nd fractions numerator: 3x^2+6x.
3rd fractions numerator:3x^2+3x.

anonymous · Answer

Since the denominators are all the same the numerators combine and are all over x(x+1)(x+2).
Then I simplified the numerators...

   3x^2+9x+6 
+ 3x^2+6x
   9x^2+15x+6
-  3x^2+3x
= 3x^2+12x+6

The final answer is 3x^2+12x+6/x(x+1)(x+2).

The original Problem: 3/x + 3/x+1 - 3/x+2

anonymous · Answer

I see one error. You forgot to distribute the negative on the third fraction. You should end up with:

anonymous · Answer

3x^2+9x+6 + 3x^2+6x + -3x^2 - 3x = 3x^2 + 12x + 6

anonymous · Answer

Wait. I guess you wrote the work down incorrectly. You get the same answer.