Perform the indicated operation. (w^3 +64)/(4+ w)

(w^2 - 4w + 16)

Step-by-step explanation:

Note that w^3 +64 is the sum of two perfect cubes, which are (w)^3 and (4)^3.  The corresponding factors are (w + 4)(w^2 - 4w + 16).

Therefore,

(w^3 +64)/(4+ w) reduces as follows:

   (w^3 +64)/(4+ w)      (4 + w)(w^2 - 4w + 16)

= =  (w^2 - 4w + 16)

             4 + w                     4 + w

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