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Theory of equations

If x3+3x29x+c is of the form (xα)2(xβ), then c is equal to


f(x)=x3+3x29x+c is of the form (xα)2(xβ), showing that α is a double root so that f'(x) = 0 has also one root α, i.e., 3x2+6x9=0 has one root α. Hence, x2+2x3=0 or (x+3)(x1)=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

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