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!

You can just cube it, like the comments have suggested$$(2+\sqrt3)^3=8+12\sqrt3+18+3\sqrt3=26+15\sqrt3$$
If you want to start off with the left-hand side and gradually work your way to the right-hand side, the easiest way is to assume$$\sqrt[3]{26+15\sqrt3}=a+b\sqrt3$$Cube both sides via the binomial theorem and compare the coefficients.$$\begin{align*} & a^3+9ab^2=26\\ & 3a^2b+3b^3=15\end{align*}$$Solving gives you $a,b$.
