Thank you, Daniel! I found an exhaustive explanation on the $Wolfram MathWorld$ page:

Cube Root -- from Wolfram MathWorld

The schoolbook definition of the cube root of a negative number is (-x)^(1/3)=-(x^(1/3)). However, extension of the cube root into the complex plane gives a branch cut along the negative real axis for the principal value of the cube root as illustrated above. By convention, "the" (principal) cube root is therefore a complex number with positive imaginary part. As a result, the Wolfram Language and other symbolic algebra languages and programs that return results valid over the entire complex plane therefore return complex results for (-x)^(1/3). For example, in the Wolfram Language, ComplexExpand[(-1)^(1/3)] gives the result 1/2+isqrt(3)/2.

When considering a positive real number x, the Wolfram Language function CubeRoot[x], which is equivalent to Surd[x, 3], may be used to return the real cube root.

First two reasons are clear, correct and acceptable! But the last one is somewhat confusing, because cubic root of $-1$ in traditional interpretation is $-1$. So, in such interpretations solutions of this equation are: $1$ and $1$. In my opinion, this is a serious confusion for a lot of people...

Wolfram Alpha finds the same solutions:

But, I used Step-by-step solution and found solution with the well known formula in the traditional form:

After that I used other variant of Step-by-step solution procedure and found the solution in the traditional form, too:

So, it's not clear why the result is presented in "non-traditional" form?

In[2]:= {{x -> -(-1)^(1/3)}, {x -> (-1)^(2/3)}} // FullForm
Out[2]//FullForm= \!\(TagBox[StyleBox[RowBox[{"List", "[", RowBox[{RowBox[{"List", "[", RowBox[{"Rule", "[", RowBox[{"x", ",", 
RowBox[{"Times", "[", RowBox[{RowBox[{"-", "1"}], ",", RowBox[{"Power", "[", RowBox[{RowBox[{"-", "1"}], ",", 
RowBox[{"Rational", "[", RowBox[{"1", ",", "3"}], "]"}]}], "]"}]}], "]"}]}], "]"}], "]"}], ",", RowBox[{"List", "[", RowBox[{"Rule", "[", 
RowBox[{"x", ",", RowBox[{"Power", "[", RowBox[{RowBox[{"-", "1"}], ",", RowBox[{"Rational", "[", RowBox[{"2", ",", "3"}], "]"}]}], "]"}]}], "]"}], "]"}]}], "]"}],ShowSpecialCharacters->False,ShowStringCharacters->True,\NumberMarks->True], FullForm]\)

What is the reason to present solutions in an unusual "rational" form?

You don't really have to map ComplexExpand, this will gave you the same result:

ComplexExpand[x/.Solve[x^2 + x + 1 == 0]]    

$$\left\{-\frac{1}{2}-\frac{i \sqrt{3}}{2},-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right\}$$

ComplexExpand[Reduce[x^2 + x + 1 == 0]]

$$x=-\frac{1}{2}-\frac{i \sqrt{3}}{2}\lor x=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$$

Thank you, Frank! It's somewhat unusual solution presentation!