The correct answer is:  
Answer choice:  [A]:__________________________________________________________→
  "
 
 " ;  " 
{ u  ± 3 }
 ± 3 } " ;           
→  or, write as:  "
 u / (u − 3) " ; 
 {" 
u ≠ 3 "
}  AND:  
{" 
u ≠ -3 "
} ; 
__________________________________________________________Explanation:__________________________________________________________ We are asked to simplify:    

 ;  
Note that the "numerator" —which is:  "(u² + 3u)" — can be factored into:
                                                      →  " u(u + 3) " ;
And that the "denominator" —which is:  "(u² − 9)" — can be factored into:
                                                      →   "(u − 3) (u + 3)" ;
___________________________________________________________Let us rewrite as:
___________________________________________________________→    

  ;
___________________________________________________________→  We can simplify by "canceling out" BOTH the "
(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:
" 

 "  ;
→  And we have:
_________________________________________________________→  " 

 " ;   that is:  " u / (u − 3) " ;  { u

 } .
                                                                                and:  { u

 } .
→ which is:  "
Answer choice:  [A] " .
_________________________________________________________NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ; 
and if the denominator is "(u − 3)" ;  the denominator equals "0" when "u = -3" ;  as such:
"u

3" ; 
→ Note:  To solve:  "u + 3 = 0" ; 
 Subtract "3" from each side of the equation; 
                       →  " u + 3 − 3 = 0 − 3 " ; 
                       → u =  -3 (when the "denominator" equals "0") ;  
                       → As such:  " u 

 -3 " ; 
Furthermore, consider the initial (unsimplified) given expression:
→
  
 ;  
Note:  The denominator is:  "
(u²  − 9)" . 
The "denominator" cannot be "0" ; because one cannot "divide" by "0" ; 
As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________ →  " u² − 9 = 0 " ; 
 →  Add "9" to each side of the equation ; 
 →  u² − 9 + 9 = 0 + 9 ; 
 →  u² = 9 ; 
Take the square root of each side of the equation; 
 to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ; 
→ √(u²) = √9 ; 
→ 
| u 
| = 3 ; 
→  " u = 3" ; AND;  "u = -3 " ; 
We already have:  "u = -3" (a value at which the "denominator equals "0") ; 
We now have "u = 3" ; as a value at which the "denominator equals "0"); 
→ As such: " 
u
" ; "
u  -3
 -3 " ;  
or, write as:  " 
{ u  ± 3 }
 ± 3 } " .
_________________________________________________________