Răspuns:
[tex]\bigg\{-2; \ 1-\sqrt{3}i; \ 1+\sqrt{3}i\bigg\}[/tex]
Explicație pas cu pas:
[tex]x^3 + 8 = 0 \implies x^3 + 2x^2 - 2x^2 - 4x + 4x + 8 = 0\\[/tex]
[tex]\implies x^2(x + 2) - 2x(x + 2) + 4(x + 2) = 0[/tex]
[tex](x + 2)(x^2- 2x + 4) = 0[/tex]
[tex]x+2=0 \implies x_1=-2[/tex]
[tex]x^2 - 2x + 4 = 0[/tex]
[tex]\Delta = (-2)^2 - 4\cdot1\cdot4 = 4 - 16 = -12[/tex]
[tex]x_2 = \dfrac{-(-2)-\sqrt{-12} }{2} = \dfrac{2-2\sqrt{3} i}{2} = 1 - \sqrt{3}i\\[/tex]
[tex]x_3 = \dfrac{-(-2)+\sqrt{-12} }{2} = \dfrac{2+2\sqrt{3} i}{2} = 1+\sqrt{3}i\\[/tex]
