Prove this equality $$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
This is how I did it:
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4 \;\;\; |^{3}$$
$$2+11i+3\sqrt[3]{(2+11i)^2(2-11i)}+3\sqrt[3]{(2+11i)(2-11i)^2}+2-11i=64$$
$$4+3\sqrt[3]{(2+11i)(2-11i)} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=64$$
$$3\sqrt[3]{4+121} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=60$$
$$15 \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=60$$
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
Is this correct?
 A: Close. As lemon pointed out, you started by assuming what was needed to be proven. This is easily fixed, however.
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=x \;\;\; |^{3}$$
$$2+11i+3\sqrt[3]{(2+11i)^2(2-11i)}+3\sqrt[3]{(2+11i)(2-11i)^2}+2-11i=x^3$$
$$4+3\sqrt[3]{(2+11i)(2-11i)} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=x^3$$
$$3\sqrt[3]{4+121} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=x^3-4$$
$$15 \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=x^3-4$$
$$15 \cdot x=x^3-4$$
$$x^3-15x-4=0$$
$$(x-4)(x^2+4x+1)=0$$
Since both $\sqrt[3]{2+11i}$ and $\sqrt[3]{2-11i}$ have positive real parts, we know that their sum must have a positive part. Both roots of $x^2+4x+1$ are negative, as can easily be checked, so we have:
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
I'm not sure if this is the easiest way to go, but it's the closest to what you already have.
A: You can begin with
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=x$$
and find $x$.
Alternatively, let $z=2+11i$. You have to show that $\sqrt[3]z+\sqrt[3]{\bar z}=4$. Since $\sqrt[3]z$ and $\sqrt[3]{\bar z}$ are conjugate, it suffices to show that $\Re\sqrt[3] z=2$.
If you try to solve for $b$
$$(2+ib)^3=2+11i$$
you get
$$8-6b^2=2$$
and
$$12b-b^3=11$$
and both equations are satisfied if $b=1$.
Now, simply compute $(2+i)^3$ to check that $\sqrt[3]{2+11i}=2+i$.
