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