There must be an elegant solution to this

Let us hope that we can take the square root of $49 + 20\sqrt[3]{6}$ inside the ring $\mathbb{Z}[\sqrt[3]{6}]$ , and so look for a solution of the form $\begin{aligned} 49 + 20\sqrt[3]{6} & = \; \big(u\sqrt[3]{6} + v\sqrt[3]{36} - 1\big)^2 \; = \; u^2\sqrt[3]{36} + 6v^2\sqrt[3]{6} + 1 + 12uv - 2u\sqrt[3]{6} - 2v\sqrt[3]{36} \\ & = \; (12uv + 1) + 2(3v^2-u)\sqrt[3]{6} + (u^2-2v)\sqrt[3]{36} \end{aligned}$ for integers $u,v$ , so we require $12uv + 1 \; = \; 49 \hspace{1.5cm} 3v^2 - u \; = \; 10 \hspace{1.5cm} u^2 = 2v$ Thus $uv=4$ , and so $u^3 = 2uv = 8$ , so that $u=v=2$ . Thus $\sqrt[3]{a} + \sqrt[3]{b} \; = \; 2\sqrt[3]{6} + 2\sqrt[3]{36} \; = \; \sqrt[3]{48} + \sqrt[3]{288}$ making the answer $48 + 288 = \boxed{336}$ .

I can suggest a solution. Its a little trial and error based.

If we expand the square it gets us terms consisting [ ( a^2/3 + b^2/3) - 2× ( a^1/3 + b^ 1/3) + { 1+ 2×(ab)^1/3}] . which equals an integer + an irrational part.

Now observe that if 1/3 powers individually gets to be integer then all other must be integer. Likewise if 2/3 powers individually are integers then all other parts are integer. So only way is [2× (ab)^1/3 + 1 ] is integer and equals to 49. That gives us ab = 24^3.

Now again if 24^3 is 4^3×6^3 . But as the irrational term consists only 6 under root so 4^3 is either divided into (1) 4^3 × 1 or (2) 2^3 × 2^3 and 6^3 as 6^2 × 6 in both cases But some trials prove case 1 not possible. So case 2 is the solution.