Using Vieta's Formulas in A Cubic Equation

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Damn .. I probably was drowning to some very complex ideas to get the initial step and meanwhile you did this and I was dead for a few seconds...
Superb

arkajitbhowmik
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You could have done it with the help of the identity a³+b³+c³=3abc if a+b+c=0

popatmuhammed
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Решение правильное, но объяснение плохое. Коэффициент при X^2 равен нулю, следовательно X1+X2+X3=0.

qyreyxd