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
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$
 A: 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.
A: 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. ;)
A: 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.
A: 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
A: Here are some remarks complementing lab bhattacharjee's answer. 
The general problem is as follows : find the “first component” of
a third root of $z=a+b\sqrt{D}$, where $a,b,D$ are integers and $D$
is a non-square (when $D<0$, this “first component” will simply be the real
part).  
Now, the generic heavy algebraic machinery tells us that this “first component”
(call it $x$) is annulated by the polynomial
$$
P=64x^9 - 48ax^6 + (-15a^2 + 27b^2D)x^3 - a^3
$$ 
and that $P$ is irreducible in general, for “almost all” values of $(a,b,D)$.
The OP’s example is special in this regard because the norm $a^2-Db^2$ is a perfect
cube. But this condition alone cannot guarantee that $P$ can be factorized ; it will
again be irreducible for  “almost all” values of $(a,b,D)$ satisfying this additional condition.
So the question now becomes : what is it about the values in the OP that allows
us to find a rational value in the end ? (this corresponds to the “lucky” factorization by $x-7$ in lab bhattacharjee’s second solution, which seems to come out of nowhere). Is there just a finite set of exceptional
points $(a,b,D)$ having this property, or do they form a curve ?
A: The square of this is $2x+2\sqrt{x^2-y^2}$.  One could find this.
Regarding $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$, the modulus of this number $35+18\sqrt{-3}$ is 2197, which is $13^3$.  This means that we are looking for a number whose cube is $35+18\sqrt{-3}$.  An example of such a number is $3.5 + 0.5\sqrt{-3}$, which we cube to get the expression of $35+18\sqrt{-3}$.  
The conjucate gives $3.5-0.5\sqrt{-3}$ gives a cube of $35-18\sqrt{-3}$.
The process here is to multiply the two numbers together, to establish the modulus.  Here, the product gives $2197$, a known cube.  In general, it needs to be a cube to fall on an integer real.  One could also divide the two numbers, in the form $\sqrt[3]{35 + 18i\sqrt{3}} / \sqrt[3]{35 - 18i\sqrt{3}}$, to get a value on the unit circle.  
However, it is easier to trace around a circle of radius $\sqrt{13}$, by looking for solutions to $x^2 + 3y^2 = 52$ for solutions of the type $(x+y\sqrt{-3})/2$.  One finds numbers like $7,3$, and then $5,3$ and $2,4$, all of which are likely suspects.  
It's a similar method that i have employed in more complex integer systems.
A: In polar form, 
$$x = 35+18 i \sqrt{3} = \sqrt{13}^3 \cdot \exp\left( i \cdot \rm{atan} \frac{18\sqrt{3}}{35}\right)$$
so that
$$\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \; \cos\left(\frac{1}{3} \rm{atan} \frac{18\sqrt{3}}{35} \right)$$
which magically evaluates to $7$ in a calculator.
(I know, I know: It's not appropriate to call any one value the cube root of another. Be that as it may ...)
To verify that the calculation is exact, we'll seek $T = \tan\theta$ such that $\tan 3\theta = 18\sqrt{3}/35$.
Invoking the triple-angle tangent formula, we have
$$\tan3\theta = \frac{T ( 3 - T^2 )}{1 - 3 T^2} = \frac{18\sqrt{3}}{35}$$
If I were convinced that $T$ is a relatively nice number, I'd be willing to believe that the $T$ in the left-hand numerator accounts for the $\sqrt{3}$ in the right-hand numerator. Write $T := S \sqrt{3}$, so that:
$$\frac{S \; ( 1 - S^2 )}{1 - 9 S^2 } = \frac{6}{35}$$
Now, note that (negative) $35$ is just one away from (negative) $36$, a perfect square (and multiple of $9$, to boot), so that comparing denominators suggests $S = \pm 2$. By happy coincidence, $S=2$ gives precisely the numerator we'd need.
Therefore, $\tan\theta = T = S\sqrt{3} = 2 \sqrt{3}$, so that $\cos\theta = \frac{1}{\sqrt{1 + T^2}} = \frac{1}{\sqrt{13}}$. This gives
$$\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \cos\theta = 2$$
and we're don--- hey, waydaminnit ... $2\neq 7$! What happened?
Well, notice that we actually solved
$$\frac{S(1-S^2)}{1-9S^2} = \frac{-6}{-35} = \tan(3\theta+\pi)$$
("Wait, what's the difference?" "Just keep reading.") so that we actually got
$$T = \tan(\theta+\pi/3) = 2\sqrt{3}$$
That is, what we took for $\theta$ was in fact $\theta+\pi/3$. We can get at the $\theta$ we wanted thusly:
$$\tan\theta = \tan\left((\theta+\pi/3)-\pi/3\right) = \frac{T-\tan\frac{\pi}{3}}{1+T\tan\frac{\pi}{3}} = \frac{2\sqrt{3}-\sqrt{3}}{1+2\sqrt{3}\cdot\sqrt{3}} = \frac{\sqrt{3}}{7}$$
so that $\cos\theta = \frac{7}{2\sqrt{13}}$ and therefore $\sqrt[3]{x} + \sqrt[3]{\overline{x}} = 2 \sqrt{13} \cos\theta = 7$, as expected.

So, why the $2$ before?
When we solved using $\frac{-6}{-35}$ instead of $\frac{6}{35}$, we implicitly replaced $x$ ---a complex number having positive real and imaginary components--- with $-x$. (Recall that $\tan 3\theta$ is the ratio of these components. It can't distinguish $x$ from $-x$.) Consequently, the $T$ we first obtained corresponds to the relation:
$$\sqrt[3]{-x} = \sqrt{13} \left( \frac{1}{\sqrt{13}} + i \frac{2\sqrt{3}} {\sqrt{13}} \right) = 1 + 2 i \sqrt{3}$$
which is to say: $(1+2i\sqrt{3})^3 = - \left( 35 + 18 i \sqrt{3}\right)$.
Then, since the arguments of $x$ and $-x$ differ by $\pi$, the arguments of the calculated cube roots must differ by $\pi/3$, and our second computation gives $\theta$ corresponding to this result:
$$\sqrt[3]{x} = \sqrt{13} \left( \frac{7}{2\sqrt{13}} + i \frac{\sqrt{3}}{2\sqrt{13}}\right) = \frac{1}{2}\left( 7 + i \sqrt{3} \right)$$
