Checking that $2+\sqrt3$ is a cube root of $26 + 15 \sqrt3$

Question

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!

Answer 1

I would try to find a cubic root of $26+15\sqrt3$ of the form $a+b\sqrt3$. Since$$(a+b\sqrt3)^3=a^3+9ab^2+(3ab+3b^3)\sqrt3,$$I would try to find numbers $a$ and $b$ such that$$\left\{\begin{array}{l}a^3+9ab^2=26\\3a^2b+3b^2=15\iff(a^2+b)b=5.\end{array}\right.$$Of course, I would try first to find integer solutions. But then$$(a^2+b)b=5\implies a=2\wedge b=1.$$And it turns out that $a=2$ and $b=1$ is indeed a solution.

Answer 2

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

Answer 3

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 $a,b$.

Answer 4

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. $\,a = 2 + \sqrt{3}\,$, $\,b = 2 - \sqrt{3}\,$.