<s>[INST] Math Algebra Question with Square Roots 
For $a\ge \frac{1}{8}$, we define,
  $$g(a)=\sqrt[\Large3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[\Large3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$$ 
Find the maximum value of $g(a)$.

I came across this question in a Math Olympiad Competition and I am not sure how to solve it. Can anyone help? Thanks.
 [/INST] Hint: Let $u =\sqrt{\frac{8a-1}{3}}$ which implies $a = \frac{3u^2+1}{8}$. Then you have to find the max of \begin{align*}
\sqrt[3]{\frac{3u^2+1+u^3+3u}{8}}+\sqrt[3]{\frac{3u^2+1-u^3-3u}{8}}.
\end{align*}
Observe the expansions of $(1+u)^3$ and $(1-u)^3$.
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