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That's right. This video clip presented an operation consisting of a sum of 2 cubic roots of real numbers. The clip showed that the result must satisfy a cubic equation, but this does not imply that every one of the cubic roots is a result of the sum, which must be single valued (and real).
You are absolutely correct. If they want to consider complex solutions then complex values of ab shouldn’t be overlooked. The other option is to consider all real values only.
@@ZoidVERSE ab = 3√-1 (ab)^3 = -1 (ab)^3 + 1^3 = 0 (ab + 1) (a^2 * b^2 - ab + 1) = 0 So you get ab + 1 = 0 and (a^2 * b^2 - ab + 1) = 0 The answer from the linear root is -1 but there is still a quadratic root. After factorizing, you will get 2 more roots, which are complex.
literally today I was interested in Cardano’s theorem for solving an equation of degree 3 and for myself I composed the equation x³+3x-4=0 The discriminant turned out to be D= q²/4+p³/27=4+1=5 hence x= (2+√5)^(1/3) +(2-√5)^(1/3) I tried for a very long time to bring this to 1, but I couldn’t do it myself! Thank you for the video. (compiled with the help of a translator, I apologize for any inaccuracies)
Из чего следует, что i - вещественное число. И даже более того, i - это сразу два разных вещественных числа. 🤓 Используя полученные результаты, также можно легко доказать, что i = 0 и что 0 = 1.
