Checking that $2+\sqrt3$ is a cube root of $26 + 15 \sqrt3$ I am trying to show that $$\sqrt[3]{26 + 15 \sqrt{3}} = 2 + \sqrt{3}$$
My idea is to find the cube roots of $z=26 + 15\sqrt{3}$ via De Moivre's formula.
So $r=\sqrt{26^2 + (15\sqrt{3})^2} = \sqrt{1351}$, and $\theta = \tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)$.
Thus, $$z^{1/3} = (1351)^{2/3} \left(\cos \left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)}{3}+ \frac{2k\pi}{3} \right] + i\sin\left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26}  \big)}{3} + \frac{2k\pi} {3}\right]\right)$$ for $k = 0,1,2.$
This does not seem like I am heading in the right direction.  Any clues or hints would be greatly appreciated.  Thank you!

$$\sqrt[3]{26+15\sqrt3}=\sqrt[3]{8+3\cdot4\cdot\sqrt3+3\cdot2\cdot3+3\sqrt3}=\sqrt[3]{(2+\sqrt3)^3}=2+\sqrt3.$$
