Respuesta :

[tex]x^3+27=0 \\x^3+3^3=0 \\(x+3)(x^2-3x+9)=0 \\x^2-3x+9=0 \\a=1,b=-3,c=9 \\ \\x_{1,2}= \frac{-b\pm \sqrt{b^2-4ac} }{2a} = \frac{-(-3)\pm \sqrt{(-3)^2-4\times1\times9} }{2\times1} =\frac{3\pm \sqrt{-27} }{2} =\frac{3\pm 3\sqrt{3}i }{2} \\ \\(x+3)(x-\frac{3+ 3\sqrt{3}i }{2})(x-\frac{3- 3\sqrt{3}i }{2})=0 [/tex]
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