Determine sum of all complex solutions of $x^3+3=0, \ x^4+4=0, \ x^5+5=0$ 
Determine sum of all complex solutions of $x^3+3=0, \ x^4+4=0, \ x^5+5=0$

This problem is confusing me a lot. Any idea where my logic is wrong?
$$x^3+3 =0\iff x^3 = -3 \Rightarrow x_1=(-3)^\frac{1}{3} \text{ for } x_1\in\mathbb{R} \\ x_1 \text{ is the only real solution, and the remaining two are complex.} \\ \text{From Viete's relations: } x_1+x_2+x_3=0 \Rightarrow x_2+x_3=-x_1=3^\frac{1}{3} \\ x^4 = -4 \text{ has no real solutions, so the sum of its complex solutions is } 0 \text{ (Again, Viete)} \\ \text{For } x^5 =-5, \text{ the sum of its complex roots, applying the same logic from the first equation, is } 5^\frac{1}{5} \\ \text{Thus, the sum of all complex roots is } 3^\frac{1}{3} + 5^\frac{1}{5}.$$
However, my workbook says the sum is $0$, but gives no explanation. Any clues?

Another way to see that the sum of the roots are $0$:
Consider $$(x^3+3)(x^4+4)(x^5+5)=0$$ Expanding gives a polynomial with no $x$-term, hence by Vietè's formulas the sum of the roots are $0$.
