Question 199846
great  problem
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to  factor,,,,( x^3 +216),,,,first   examine  the  last  term  for  factors,  1,216, 2,108,3, 72, 4, 54, 6,36
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now  we  try  to find  which  factor  will  go  evenly  into  the  function.,,,,usually  we  do  this  by  division,  either  polynomial  or  synthetic.
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using  synthetic  division  on  a  best  quess  of  (x+6) factor  or  x=-6  zero:
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-6,,,,,,,,,,,,,,1     0      0      216,,,,,,,,,,,,,,,,,,remember  to  fill  for  x^2  and  x terms
,,,,,,,,,,,,,,,,,,,-6    36    -216
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,,,,,,,,,,,,,,1,,,-6,,,36,,,,,,,,0,,,,,,,,,,,,,,,as  remainder  is  zero,  this  is  a  factor
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(x+6) ( x^2 -6x +36),,,,,,,,,,,to  factor  the  second  term,  use  the  quadratic  formula
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a= (1),,,,b=(-6) ,,,,,,c=(36)
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x =[ -(-6) +/- { (-6)^2 - 4 (1) (36) } ^(1/2) ] / 2 ( 1)
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x= [ 6 +/- { 36 -144)^(1/2) } ] /2
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x= [ 6 +/- {-108)^(1/2) } ] /2,,,with  sqrt (-108)  = sqrt (-1) * sqrt (108 = 10.4 i
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x= 3 +/- ( 5.2 i ),  x= 3+5.2i ,,,,,or  x = 3-5.2i
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Collecting  all  the  zeros,,,,x= (-6),,,, ( 3+5.2i) ,,,,,(3-5.2i)
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checking  
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( x+6) ( x-(3+5.2i) )  ( x-(3-5.2i) )
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(x+6) ( x-3-5.2i) ( x-3+5.2i),,,,,good  summary  of  all  factors
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(x+6) ( x^2 -6x +36)
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x^3 + 216,,,,,ok