#### Answer

$x = 1.36+1.26~i$
$x = -1.775+0.546~i$
$x = 0.414-1.810~i$

#### Work Step by Step

$x^3+4-5i = 0$
$x^3 = -4+5i$
We can find $r$:
$r = \sqrt{(-4)^2+(5)^2} = \sqrt{41} = 6.403$
We can find the magnitude of $x$:
$r^{1/3} = \sqrt[3] {6.403} = 1.857$
Note that $~~-4+5i~~$ is in the second quadrant:
$tan~\theta = -\frac{5}{4}$
$\theta = tan^{-1}(-\frac{5}{4})+180^{\circ}$
$\theta = 128.66^{\circ}$
Then:
$\frac{128.65^{\circ}+360^{\circ}~k}{3} = 42.89^{\circ}+120^{\circ}~k$
We can find the solutions for $x$:
When $k = 0$:
$x = 1.857[cos~(42.89^{\circ})+i~sin~(42.89^{\circ})]$
$x = 1.36+1.26~i$
When $k=1$:
$x = 1.857[cos~(162.89^{\circ})+i~sin~(162.89^{\circ})]$
$x = -1.775+0.546~i$
When $k=2$:
$x = 1.857[cos~(282.89^{\circ})+i~sin~(282.89^{\circ})]$
$x = 0.414-1.810~i$