If  x3+3x2−9x+c  is of the form (x−α)2(x−β),then c is equal to

f(x)=x3+3x2−9x+c is of the form (x−α)2(x−β), showing that a is a double root so that f′(x)=0 has also one root d,i.e., 3x2+6x−9=0 has one root α. Hence,x2+2x−3=0 or (x+3)(x−1)=0 has the root α which can be either -3 or 1.If α= 1, then f (x)=0 gives c- 5 = 0 or c=5.If α= -3, then f (x) = 0 gives −27+27+27+c=0∴ c=−27