How does one evaluate $\sqrt[3]{x + iy} + \sqrt[3]{x - iy}$?

As I said in a previous answer, finding out that such a simplification occurs is exactly as hard as finding out that $35+18\sqrt{-3}$ has a cube root in $\Bbb Q(\sqrt{-3})$ (well actually, $\Bbb Z[(1+\sqrt{-3})/2]$ because $35+18\sqrt{-3}$ is an algebraic integer).

Suppose $p,q,d$ are integers. Then $p+q\sqrt d$ is an algebraic integer, and so is its cube root, and we can look for cube roots of the form $(x+y\sqrt d)$ with $2x,2y \in \Bbb Z$

So here is what you can do :

• compute analytically the $9$ possible values of $(p+q\sqrt{d})^{1/3}+(p-q\sqrt{d})^{1/3}$, using the polar form to take cube roots, so this involves the use of transcendental functions like $\arctan, \log, \exp$ and $\cos$. Does one of them look like an integer $x$ ? if so, let $y = 3qx/(x^3+p)$ and check algebraically if $(x+y\sqrt d)^3 = 8(p+q\sqrt d)$.

• try to find a semi-integer root to the degree $9$ polynomial $64x^9-48px^6+(27dq^2-15p^2)x^3-p^3 = 0$. It will necessarily be of the form $\pm z$ or $\pm z/2$ where $z$ is a divisor of $p$, so you will need to decompose $p$ into its prime factors.

• see if the norm of $p+q\sqrt d$ is a cube (in $\Bbb Z$). If it's not, you can stop. If you find that $p^2-dq^2 = r^3$ for some integer $r$, then try to find a semi-integer root to the degree $3$ polynomial $4x^3-3rx-p = 0$. Again you only need to check divisors of $p$ and their halves. The check that the norm was a cube allows you to make a simpler computation.

• study the factorisation of $p+q\sqrt d$ in the ring of integers of $\Bbb Q(\sqrt d)$. This involves again checking that $p^2-dq^2$ is a cube $r^3$ (in $\Bbb Z$), then find the factorisation of the principal ideal $(p+q\sqrt d)$ in prime ideals. To do this you have to find square roots of $d$ modulo some prime factors of $r$ and find the relevant prime ideals. If $(p+q\sqrt d)$ is the cube of an ideal, you will need to check if that ideal is principal, say $(z)$ (so you may want to compute the ideal class group of $\Bbb Q(\sqrt d))$, and then finally compute the group of units to check if $(p+q\sqrt d)/z^3$ is the cube of a unit or not (especially when $d>0$, where that group is infinite and you need to do some continued fraction computation. otherwise it's finite and easy in comparison).

I would say that methods $2$ and $4$ are strictly more complicated than method $3$, that checking if a polynomial has some integer roots is not too much more evil than doing a prime factorisation to check if some integer is a cube or not. And if you can bear with it, go for method $1$, which is the best, and you don't need a high precision.

This is unavoidable. If you use Cardan's formula for the polynomial of the method $3$ you will end up precisely with the expression you started with, and that polynomial is the one you started with before considering simplifying that expression.
Finally, I think what you want to hear is There is no formula that allows you to solve a cubic real polynomial by only taking roots of real numbers.

Consider $a^3 =35+ 18i\sqrt{3}$ and $b^3=35- 18 i\sqrt{3}$

You are supposed to find $a+b$ $(a+b)^3=a^3+b^3+3ab(a+b) \implies(a+b)^3-3ab(a+b)=a^3+b^3$

Let $a+b=x \implies x^3-3ab(x)=70$ and $ab= [(35-18i\sqrt3)(35+18i\sqrt3)]^{1/3}=13$

You get $x^3-39x-70 =0$, product of the roots is $70$ and sum of the roots is $0$. You can do clever factorization or guess the roots(It has only 8 divisors). SO, number of possible should be just $8 \choose 3$. I kinda flicked your trick. ;)

As $(a+ib)^{\frac13}$ is a multi-valued function, it is technically correct to state that one of the three values of $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}}$ is $7$

Method $1:$

Let $\sqrt[3]{35 + 18i\sqrt{3}}=a+ib\iff \sqrt[3]{35 - 18i\sqrt{3}}=a-ib$

So, $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 2a$

As this needs to be $7,$ let's try with $a=\frac72$

$$\text{So, }\left(\frac72+ib\right)^3=35+18i\sqrt3 $$

$$\implies (7+2i b)^3=8(35+18i\sqrt3)$$

$$\implies 7^3+3\cdot7^2\cdot2ib +3\cdot 7\cdot(2ib)^2+(2ib)^3=280+144i\sqrt3$$

Equating the real parts, $343-84b^2=280\implies 4b^2=3\implies b=\pm\frac{\sqrt3}2$

Equating the imaginary parts, $294b-8b^3=144\sqrt3$

$\implies 2b(147-4b^2)=144\sqrt3\implies 2b=\sqrt3\implies b=\frac{\sqrt3}2$

As $\sqrt[3]{35 + 18i\sqrt{3}}=\frac{7+\sqrt3i}2,\sqrt[3]{35 - 18i\sqrt{3}}=\frac{7-\sqrt3i}2$

Again, $\sqrt[3]{35 + 18i\sqrt{3}}=\sqrt[3]{35 + 18i\sqrt{3}}\cdot \sqrt[3]1=\frac{7+\sqrt3i}2\cdot w$ where $w$ is a cube root of $1$

As the cube root of $1$ are $1,\frac{-1\pm\sqrt3i}2$

So, one of the other values of $\sqrt[3]{35 + 18i\sqrt{3}}$ is $\left(\frac{7+\sqrt3i}2\right)\cdot\left(\frac{-1+\sqrt3i}2\right)=\frac{-5+3\sqrt3i}2$

The corresponding value of $\sqrt[3]{35 - 18i\sqrt{3}}$ is $\frac{-5-3\sqrt3i}2$

Similarly, another value of $\sqrt[3]{35 + 18i\sqrt{3}}$ is $\left(\frac{7+\sqrt3i}2\right)\cdot\left(\frac{-1-\sqrt3i}2\right)=-1+2\sqrt3i$

The corresponding value of $\sqrt[3]{35 - 18i\sqrt{3}}$ is $-1-2\sqrt3i$

Method $2:$

Let $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = a$

$$\implies a^3=35 + 18i\sqrt{3}+35 - 18i\sqrt{3}-3\sqrt[3]{35 + 18i\sqrt{3}}\sqrt[3]{35 - 18i\sqrt{3}}a$$

$$\implies a^3-39a-70=0$$ as $(35-18i\sqrt3)(35+18i\sqrt3)=2197=13^3$

$$\implies a^3-7^3-39(a-7)=0$$ $$\implies (a-7)(a^2+7a+7^2-39)=0$$ $$\implies(a-7)(a+2)(a+5)=0 $$

Here's one way for square roots:

$$(\sqrt{x+y}+\sqrt{x-y})^2 = 2x + 2\sqrt{x^2-y^2}.$$

Perhaps this is easier to evaluate. For cube roots:

$$ (\sqrt[3]{x+y}+\sqrt[3]{x-y})^3 = 2x + 3\sqrt[3]{(x+y)(x^2-y^2)}+3\sqrt[3]{(x-y)(x^2-y^2)}. $$

Of course, to get the coefficients in all of these I am using the binomial theorem. There really is no guarantee that these are easier to deal with in general.