Your calculations are correct, but it is necessary to complete the Cardano's method. Once you have calculated $a$ and $b$, the roots of the depressed cubic are as follows:
$$
\displaystyle z_{1}=a+b \\
{\displaystyle z_{2}=a\cdot \left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)+b\cdot \left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)} \\
{\displaystyle z_{3}=a\cdot \left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)+b\cdot \left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)}
$$
Since in your case $a=1+2/\sqrt{3}$ and $b=1-2/\sqrt{3}$ (see below for the denesting procedure to obtain these values), the formulas give
$$z_1=2 \\
z_2=-1+2 i \\
z_3=-1-2 i$$
As $x=z+1$, you have
$$x_1=3 \\
x_2=2 i \\
x_3=-2 i$$
EDIT: as correctly stated in the comments, a key issue in applying Cardano's method is that, in some cases, there is the need to denest some cubic roots. This may sometimes be quite difficult. Some methods have been previously reported in the links provided in one of the comments. I would suggest a possible approach that sometimes works well for the radicand of the form $J+K\sqrt{n}$. The methods includes these steps:
set the cube root in the form $\sqrt[3]{J\pm K\sqrt{n}}$, with $J$ and $K$ integers;
assume that the radicand $A=J\pm K\sqrt{n}$ can be expressed as $(j\pm k\sqrt{n})^3$, with $j$ and $k$ rational numbers;
after expanding $(j\pm k\sqrt{n})^3$ and dividing its terms in two groups whose sums are equal to $J$ and $K\sqrt{n}$, use the resulting equations to determine $j/k$. This is the longer step, as it requires to search the rational roots of a new cubic equation using the rational root theorem, which sometimes may be tricky;
finally, determine the values of $j$ and $k$.
To better illustrate this method, let us try it for the specific case $\sqrt[3]{5+ \frac{26\sqrt{3}}{9}}$ (the same method can be used for the case where the radicand is $5-\frac{26\sqrt{3}}{9}$). Firstly, we have to set the radicand so that $J$ and $K$ are integers:
$$\sqrt[3]{5 + \frac{26\sqrt{3}}{9}}=\frac{1}{3} \sqrt[3]{135+ 78\sqrt{3}} = \frac{1}{3} \sqrt[3]{A} $$
Now let us hypothesize $A=(j+k\sqrt{3})^3$. Therefore
$$A= j^3+3\sqrt{3}j^2k+ 9jk^2+3\sqrt{3}k^3\\
=j(j^2+9k^2)+3k(j^2+k^2)\sqrt{3}$$
so that we can write
$$j(j^2+9k^2)=135\\
3k(j^2+k^2)=78$$
Note that $j$ and $k$ have to be both positive. From the two equations above we have
$$78\cdot j(j^2+9k^2) =135\cdot
3k(j^2+k^2)$$
We have now to try to determine $j/k$. Dividing both members to $k^3$ and moving all terms to the LHS, we have
$$78\left(\frac{j}{k}\right)^3
- 405 \left(\frac{j}{k}\right)^2 + 702\left(\frac{j}{k}\right) - 405=0 $$
Setting $x=j/k$ and simplifying the coefficients, we get
$$26 x^3-135 x^2+234x-135=0$$
Using the rational root theorem, we can search for a rational root $p/q$ for the last equation, where the integer $p$ divides $135=3^3\cdot 5$ and the integer $q$ divides $26=2\cdot 13$. To speed up the search of a real root, it can be observed that for $x=1$ and $x=2$ the LHS gives $-10$ and $1$, respectively, so that the value of one real root has to be between $1$ and $2$. After few trials, we easily get $x=3/2$. The equation can then be rewritten as
$$\left(x-\frac 32\right)\left( 26x^2-96x+90\right)=0$$
from which we directly get that the other two roots are not real.
Since $x=j/k=3/2$, we can finally determine $j$ and $k$ by making the substitution $k=2j/3$ in the initial equations. For example, substituting in the equation $(j^2+9k^2)=135$, we have
$$j\left[j^2+9\left(\frac{2j}{3}\right)^2\right]=135$$
$$5j^3=135$$
and reminding that $j$ and $k$ are positive,
$$j=3$$
$$k=2$$
We can now conclude that
$$A=(3+2\sqrt{3})^3$$
so that the initial cubic root is
$$\sqrt[3]{5 + \frac{26\sqrt{3}}{9}}=\frac 13 \sqrt[3]{A}= \frac 13 \left(3+2\sqrt{3}\right)\\=1+\frac{2}{\sqrt{3}}$$
Again, it must be pointed out that this method works only in some cases (even when the rational $j$ and $k$ exist, the most important limiting step is the search of the rational root $x$, which as already stated can be very difficult).
