If the equation x3 3x2 m 0 has roots that form an arithmet

Solution

in an arithmetic sequence a,b,c b = (a+c)/2 or a+c = 2b b-a = c-b x^3 +3x^2 -m = 0 we can write (x-a)(x-b)(x-c) = x^3 +3x^2 -m x^3-(a+b+c)x^2+(ab+ac+bc)x-abc = x^3 + 3 x^2 -m compare like terms -a-b-c = 3 and -m = -abc and ab+ac+bc = 0 a+b+c =-3 2b + b = -3 3b =-3 b = -1 from ab+bc+ca = 0 b(a+c)+ca = 0 b(2b)+ca = 0 2b^2 +ca = 0 2* -1^2 +ca = 0 ca = -2 from -m = -acb m = acb = (-2)(-1) = 2 so m = 2