Respuesta :
Answer: A. 2
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Short explanation:
Plug z = 2 into the equation and you'll find that
z^3 + 8 = 2^3 + 8 = 8+8 = 16
but the result should be 0 since the original equation has 0 on the right side. Therefore, z = 2 is not a solution.
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Longer explanation:
Solve the equation for z to find the three roots.
z^3 + 8 = 0
(z+2)(z^2 - 2z + 4) = 0 ... sum of cubes factoring rule
z+2 = 0 or z^2 - 2z + 4 = 0
z = -2 or z^2 - 2z + 4 = 0
We see that z = -2 is one root. To find the other two roots, use the quadratic formula to solve z^2 - 2z + 4 = 0 for z
[tex]z = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\z = \frac{-(-2)\pm\sqrt{(-2)^2-4(1)(4)}}{2(1)}\\\\z = \frac{2\pm\sqrt{-12}}{2}\\\\z = \frac{2\pm\sqrt{-1*4*3}}{2}\\\\z = \frac{2\pm\sqrt{-1}*\sqrt{4}*\sqrt{3}}{2}\\\\z = \frac{2\pm i*2*\sqrt{3}}{2}\\\\z = \frac{2\pm 2i*\sqrt{3}}{2}\\\\z = \frac{2(1\pm i*\sqrt{3})}{2}\\\\z = 1\pm i\sqrt{3}\\\\z = 1+ i\sqrt{3} \ \text{ or } \ z = 1- i\sqrt{3}\\\\[/tex]
Therefore, the three roots of z^3+8=0 are [tex]z = -2, \ z = 1+i\sqrt{3}, \ z=1-i\sqrt{3}[/tex]
The value z = 2 is not part of the list of solutions.
We can verify each solution by plugging it back into the original equation. For instance, with z = -2, we get
z^3 + 8 = 0
(-2)^3+8 = 0
-8+8 = 0
0 = 0 ... solution is confirmed.
I'll let you check the other solutions.
