Question

Determinați rădăcinile de ordin III ale lui -1.​

Answer 1

 

$\displaystyle\\x^3=-1\\\\\text{Daca calculam asa: }\\\\x=\sqrt[\b3]{-1}=-1\\\\\text{nu e corect deoarece ecuatia este de gradul 3 si are 3 solutii.}\\\\\text{Rezolvare:}\\\\x^3=-1\\x^3+1=0\\x^3+1^3=0\\\text{folosim formula: }~a^3+b^3=(a+b)(a^2-ab+b^2)\\(x^3+1)=(x+1)(x^2-x\cdot1+1^2)\\(x+1)(x^2-x+1)=0\\x+1=0~~\text{sau}~~x^2-x+1=0$

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$\displaystyle\\x+1=0\\\boxed{x_1=-1}\\\\x^2-x+1=0\\\\x_{23}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\sqrt{-3}}{2}=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i  \\\\\boxed{x_2=\frac{1}{2}+\frac{\sqrt{3}}{2}i}\\\\\boxed{x_3=\frac{1}{2}-\frac{\sqrt{3}}{2}i}$