Do there exist shorter ways of solving for $x$ (or other smarter methods)?

Question

Problem:

If$$\frac 13\Bigr(\sqrt [3] {6\big(9+5\sqrt{3}\big)}+\sqrt [3] {6\big(9-5\sqrt{3}\big)}\Bigr)=2,\tag1$$ then $$\sqrt [3] {\frac x9}\bigr(\sqrt [3] {9+5\sqrt{3}}+\sqrt [3] {9-5\sqrt{3}}\bigr)=2.\tag2$$Prove Eq. $(1)$ and then solve for $x$ in Eq. $(2)$.

I made up this problem. I discovered the first formula, inspired by how $\sqrt [3]{2+\sqrt{5}}+\sqrt [3]{2-\sqrt{5}}=1$. And then I took out the six and with some trial and error, I found $x$ (because $x\div 9$ is a familiar fraction). My question is, however, how does one solve equations like these in a shorter amount of time? Are there any techniques I should know?

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My Attempt Regarding Solution for $x$:

If Eq. $(1)$ is true, then $$\sqrt [3]{6\big(9+5\sqrt{3}\big)}+\sqrt [3]{6\big(9-5\sqrt{3}\big)}=2\times 3 = 6.$$ Now I think one can do the following, but I am not too sure. $$\begin{align}6&=\sqrt [3]{6\big(9+5\sqrt{3}\big)}+\sqrt [3]{6\big(9-5\sqrt{3}\big)} \\ &= \sqrt[3]6\sqrt[3]{9+5\sqrt{3}}+\sqrt [3]6\sqrt[3]{9-5\sqrt3} \\ &= \sqrt[3]6\bigr(\sqrt[3]{9+5\sqrt 3}+\sqrt[3]{9-5\sqrt 3}\bigr).\end{align}$$ It follows, then, that $$\begin{align}6&=\sqrt [3] 6\bigg(\frac{2}{\sqrt [3] x\div \sqrt [3] 9}\bigg) \\ &=2\sqrt[3]6 \sqrt[3]{\frac 9x}.\end{align}\tag{$\ast$}$$ Now notice that $6=\sqrt [3] 6^3$ thus $6\div \sqrt[3]6=\sqrt[3]6^2=\sqrt[3]{6^2}=\sqrt[3]{36}$. $$\therefore \sqrt [3] {36}=2\sqrt [3]{\frac 9x}\tag{$\ast\ast$}$$ which from $(\ast)$ is equivalent to the following: $$\sqrt [3]{4x}=2\tag*{$\because 36\div 9=4$}.$$ Now that I only have one cube, I can finally get rid of it by cubing both sides (because I know that $(\sqrt[3]a+\sqrt[3]b)^3$ does not get rid of all the cubes), id est $$\begin{align}4x&=2^3=8 \\ \therefore x&=8\div 4 = 2.\end{align}$$

I believe in both equations $(\ast)$ and $(\ast\ast)$, I could have cubed both sides, but I wanted to use as minimal steps as I could, so I tried nailing down to just one simple cube root before cubing the equation.

What I do not believe is that I proved Eq $(1)$. I do not think I proved it... but solving for $x$ to find that $x=2$, if we substitute that back in, it follows that Eq. $(2)$ is correct and thus Eq. $(1)$ is correct. But does this actually prove Eq. $(1)$? What if I was unaware of Eq. $(2)$? My attempt is as follows:

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My Attempt Regarding Proof of Eq. $(1)$:

I got that $$\begin{align} 36&= \big(\sqrt [3]{9+5\sqrt 3}+\sqrt [3]{9-5\sqrt{3}}\big)^3 \\ &= 9+5\sqrt{3}+9-5\sqrt{3}+3\sqrt[3]{(9+5\sqrt{3})}\sqrt [3]{(9-5\sqrt{3})}\bigr(\sqrt [3]{9+5\sqrt 3}+\sqrt [3]{9-5\sqrt{3}}\bigr) \\ &= 9\require{\cancel}\cancel{+\,5\sqrt{3}}+9\cancel{-\,5\sqrt{3}}+3\sqrt[3]{(9+5\sqrt{3})(9-5\sqrt{3})}\sqrt[3] {36} \\ &= 9+9+3\sqrt [3]{36(9+5\sqrt 3)(9-5\sqrt 3)}\\ &= 18+3\sqrt [3]{36\bigr(9^2-(5\sqrt 3)^2\bigr)}\qquad\qquad\qquad\qquad\qquad\binom{\because (a+b)(a-b)}{= a^2-b^2}\\ &= 18+3\sqrt [3]{36(81-25\times 3)} \\ &= 18+3\sqrt[3]{36(81-75)} \\ &= 18+3\sqrt [3]{36\times 6} \\ &= 18+3\times 6\qquad\qquad\qquad\qquad\qquad\bigr(\because 36=6^2\Rightarrow 36\times 6 = 6^2\times 6 = 6^3\bigr) \\ &= 18 + 18 \\ &= 36.\;\color{green}{\checkmark}\end{align}$$ I used the formula $(a+b)^3=a^3+b^3+3ab(a+b)$ and substituted $a=\sqrt[3]{9+5\sqrt{3}}$ and $b=\sqrt[3]{9-5\sqrt{3}}$. Then, I simplified (id est combined like radicals, cancelled out terms, etcetera).

I now believe I did just fine, but this was tiring. Is there a shorter way of solving such equations? Are there any smarter ways of carrying out these attempts of mine?

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Thank you in advance (and apoligies for the long post).

Answer 1

Let $a = \sqrt[3]{6(9+5\sqrt{3})}$ and $b=\sqrt[3]{6(9-5\sqrt{3})}$ then:

$$\require{cancel} a^3+b^3 = 6\cdot\left(9+\bcancel{5\sqrt{3}}\right) + 6 \cdot\left(9-\bcancel{5\sqrt{3}}\right) = 108 \\[10px] a b = \sqrt[3]{6^2 \cdot\left(9+5\sqrt{3}\right)\cdot\left(9-5\sqrt{3}\right)} = \sqrt[3]{6^2\cdot(9^2 - 5^2 \cdot 3)} = \sqrt[3]{6^2\cdot 6} = 6 $$

It follows that $\,(a+b)^3 = a^3+b^3+3ab(a+b)=108+18(a+b)\,$, so $\,t = a+b\,$ is a root of $\,t^3 - 18t - 108=0\,$. The latter cubic has the easy-to-find rational root $\,t=6\,$, and it is straightforward to show that that's the only real root, which concludes the proof of $(1)$.

Answer 2

Let \begin{eqnarray*} y=\sqrt [3] {6\big(9+5\sqrt{3}\big)}+\sqrt [3] {6\big(9-5\sqrt{3})}, \end{eqnarray*} cube this and we have \begin{eqnarray*} y^3=108+3y \sqrt [3] {6\big(9+5\sqrt{3}\big)}\sqrt [3] {6\big(9-5\sqrt{3})}=18y+108. \end{eqnarray*} This factorises $(y-6)(y^2+6x-18)=0$ so $y=6$, confirming the first equation.

The second equation gives \begin{eqnarray*} \sqrt [3] { \frac{x}{9}} =\frac{ \sqrt[3]{6}}{3} =\frac{\sqrt[3]{2}}{\sqrt[3]{9}} \end{eqnarray*} So we have $x=2$ as required.