megankaye4 megankaye4
  • 13-12-2018
  • Mathematics
contestada

What are the real and complex solutions of the polynomial equation X^3-216=0

Respuesta :

frika
frika frika
  • 22-12-2018

Answer:

[tex]6,\ -3+3\sqrt{3}i,\ -3-3\sqrt{3}i[/tex]

Step-by-step explanation:

Factor the equation [tex]x^3-216=0[/tex] using formula for difference of the cubes:

[tex]x^3-216=(x-6)(x^2+6x+36),[/tex]

then

[tex]x-6=0\ or\ x^2+6x+36=0.[/tex]

1. The equation [tex]x-6=0[/tex] has real solution [tex]x=6.[/tex]

2. The equation [tex]x^2+6x+36=0[/tex] has negative discriminant [tex]D=6^2-4\cdot 36=36-144=-108,[/tex] then it has two complex solutions

[tex]x_{1,2}=\dfrac{-6\pm 6\sqrt{3}i}{2}=-3\pm3\sqrt{3}i.[/tex]

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