How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate  the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$

I came across very similar problems while doing Maths Olympiad in high school and  had only one approach to it then.
Observe the property for $a,b,c \in \Bbb K$ $$ \text {if } a+b+c=0,\ \text{ then  }\  a^3+b^3+c^3=3abc \quad (\ast)$$
This can easily be proved by expanding $(a+b+c)^3$
$$\begin{array}{rll}(a+b+c)^3&=&a^3+b^3+c^3+\underbrace{3a^2b+3a^2c+3b^2c+3b^2a+3c^2a+3c^2b}_X+6abc=0\\X&=&3ab\underbrace{(a+b)}_{=-c}+3bc\underbrace{(b+c)}_{=-a}+3ca\underbrace{(c+a)}_{=-b}=-9abc \\ (a+b+c)^3&=&a^3+b^3+c^3-9abc+6abc\ =\ 0 \ \implies\  (\ast)\end{array}$$
In the problem above you have an expression of the form
$$\sqrt[3]{x+\sqrt y}+\sqrt[3]{x-\sqrt y}=z$$which can also translate to $$\underbrace{\sqrt[3]{x+\sqrt y}}_a+\underbrace{\sqrt[3]{x-\sqrt y}}_b +\underbrace{(-z)}_c=0$$
Applying the property abover
$$\begin{array}{rll}\underbrace{\left(\sqrt[3]{x+\sqrt y}\right)^3+\left(\sqrt[3]{x-\sqrt y}\right)^3}_{2x} +(-z)^3&=&3\underbrace{\left(\sqrt[3]{x+\sqrt y}\right)\left(\sqrt[3]{x-\sqrt y}\right)}_{\text{conjugates}}(-z) \\ \implies \quad 2x-z^3&=&-3z\cdot \sqrt[3]{x^2-y}\end{array}$$
In conclusion the problem resolves to finding the roots
$$z^3-3z\cdot \sqrt[3]{x^2-y} -2x = 0$$
I have also been curious about this, I asked this sometimes ago
