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

huh? im pretty sure it is in 199

step-by-step explanation:

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8