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While the method above is indeed valid, one could also solve the problem using calculus: Let P(x)=x3+3ax2+3bx+c. Then P'(x)=3x2+6ax+3b. For there to be a double root, there must exist some x such that P(x)=0 and P'(x)=0. P'(x)=0 iff 3x2+6ax+3b=0, i.e. x2+2ax+b=0. Now, P(x)=0 for the same...