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 α  is a double root so that f'(x)=0  has also one root α, 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