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Polynomials with order 3 have at most 3 solutions. Arranging the eqn in another order: x^3 - 27 = 0 x^3 - 3^3 = 0 (x - 3) (x^2 + 3x +9) = 0 Apparently, x = 3 and there are 2 more complex solutions which can be obtained by quadratic formula. DONE😊
Use Euler's formula for questions like this. (x/3)^3 = 1 = e^(2πi)=> roots e^0, e^(2πi/3), e^(4π/3). Solutions 3 x the above list so 3, 3(-1 +√3 i)/2, 3(-1 - √3i)/2.
In my simple mind, the answeres would be 3, and absolute value of 3 and -3, the remainder of the solutions seem to be changing the equation. Someone who is smarter than I am, please explain why you would care about creating a negative... this seems overly complicated for the sake of complicating it.