<s>[INST] 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?
 [/INST] 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 $\sqrt[3]{2+11i}=2+i$.
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