The correct answer is:  Answer choice:  [A]:
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→  "
 " ;  " { uÂ
 ± 3 } " ;Â
      →  or, write as:  " u / (u − 3) " ;  {" u ≠3 "}  AND:  {" u ≠-3 "} ;Â
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Explanation:
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 We are asked to simplify:
 Â
 Â
; Â
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)" ;
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Let us rewrite as:
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→  Â
 ;
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→  We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:
"Â
" Â ;
→  And we have:
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→  "
" ;  that is:  " u / (u − 3) " ;  { u
} .
                                        and:  { u
} .
→ which is:  "Answer choice:  [A] " .
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NOTE:Â Â The "denominator" cannot equal "0" ;Â since one cannot "divide by "0" ;Â
and if the denominator is "(u − 3)" ;  the denominator equals "0" when "u = -3" ;  as such:
"u
3" ;Â
→ Note:  To solve:  "u + 3 = 0" ;Â
 Subtract "3" from each side of the equation;Â
            →  " u + 3 − 3 = 0 − 3 " ;Â
            → u =  -3 (when the "denominator" equals "0") ;Â
Â
            → As such:  " uÂ
-3 " ;Â
Furthermore, consider the initial (unsimplified) given expression:
→ Â
; Â
Note:  The denominator is:  "(u²  − 9)" .Â
The "denominator" cannot be "0" ;Â because one cannot "divide" by "0" ;Â
As such, solve for values of "u" when the "denominator" equals "0" ; that is:
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→  " u² − 9 = 0 " ;Â
 →  Add "9" to each side of the equation ;Â
 →  u² − 9 + 9 = 0 + 9 ;Â
 →  u² = 9 ;Â
Take the square root of each side of the equation;Â
 to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;Â
→ √(u²) = √9 ;Â
→ | u | = 3 ;Â
→  " u = 3" ; AND;  "u = -3 " ;Â
We already have: Â "u = -3" (a value at which the "denominator equals "0") ;Â
We now have "u = 3" ; as a value at which the "denominator equals "0");Â
→ As such: " u
" ; "uÂ
-3Â " ; Â
or, write as: Â "Â { uÂ
 ± 3 } " .
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