<s>[INST] 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!
 [/INST] Let $\,a = \sqrt[3]{26 + 15 \sqrt{3}}\,$, $\,b=\sqrt[3]{26 - 15 \sqrt{3}}\,$, then $\require{cancel}\,a^3+b^3=26 + \cancel{15 \sqrt{3}} + 26 - \cancel{15 \sqrt{3}} = 52$ and $\,ab=\sqrt[3]{26^2 - 15^2 \cdot 3} = 1\,$.
It follows that $\,(a+b)^3 = a^3 + b^3 + 3ab(a+b) = 52 + 3(a+b)\,$, so $\,t=a+b\,$ satisfies the equation $\,t^3 - 3t -52 =0 \iff (t-4)(t^2+4t+13) = 0\,$ with the only real root $\,t = 4\,$.
Then $\,a+b=4\,$, $\,ab=1\,$ so $\,a,b\,$ are the roots of $\,x^2 - 4x + 1=0 \iff x = 2 \pm \sqrt{3}\,$ with $\,a\,$ being the larger one i.e. $\,a = 2 + \sqrt{3}\,$, $\,b = 2 - \sqrt{3}\,$.
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