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.
 A: 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 $(1-u)^3$.
A: Note the 
$$(a+b)^3=a^3+b^3+3ab(a+b)$$
so
$$g^3(a)=2a+3\sqrt[3]{a^2-\dfrac{(a+1)^2}{9}\cdot\dfrac{8a-1}{3}}\cdot g(a)$$
A: In general we cannot rewrite a cubic radical in a simple way. However both
radicals in $g(a)$ are of a special form that can be denested, because we
can find two cubic powers $X^3,Y^3$ such that $X, Y $ are quadratic conjugate irrationals and
\begin{equation*}
a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}=X^{3},\qquad a-\frac{a+1}{3}\sqrt{\frac{
8a-1}{3}}=Y^{3}.
\end{equation*}
If we write $x=\frac{8a-1}{3}\geq 0$, then $a=\frac{3x+1}{8}$ and $\frac{a+1
}{3}=\frac{x+3}{8}$. Consequently, the first radicand becomes 
\begin{eqnarray*}
a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}} &=&\frac{3x+1+\left( x+3\right) \sqrt{x}
}{8}=\frac{1+3\sqrt{x}+3x+\left( \sqrt{x}\right) ^{3}}{8} \\
&=&\frac{\left( 1+\sqrt{x}\right) ^{3}}{2^{3}}=X^{3},\qquad X=\frac{ 1+\sqrt{x} }{2}=\frac {1}{2}+\frac {1}{2}\sqrt {\frac {8a-1}{3}},
\end{eqnarray*}
where we used the binomial theorem for the cubic power
\begin{equation*}
\left( 1+c\right) ^{3}=1+3c+3c^{2}+c^{3},
\end{equation*}
with $c=\sqrt{x}$: 
\begin{eqnarray*}
\left( 1+\sqrt{x}\right) ^{3} &=&\left( 1+x^{1/2}\right)
^{3}=1+3x^{1/2}+3x+x^{3/2} \\
&=&1+3\sqrt{x}+3x+\left( \sqrt{x}\right) ^{3}.
\end{eqnarray*}
Similarly, using the binomial theorem for $(1-c)^3=(1-\sqrt{x})^3$, the second radical becomes
\begin{eqnarray*}
a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}} &=&\frac{3x+1-\left( x+3\right) \sqrt{x}
}{8}=\frac{ 1-3\sqrt{x}+3x-\left( \sqrt{x}\right) ^{3}}{2^{3}}  \\
&=&\frac{\left( 1-\sqrt{x}\right) ^{3}}{2^{3}}=Y^{3},\qquad Y=\frac{ 1-\sqrt{x} }{2}=\frac {1}{2}-\frac {1}{2}\sqrt {\frac {8a-1}{3}}.
\end{eqnarray*}
Now we find easily that for $a\geq 1/8$ the function $g(a)$ is 
constant
\begin{equation*}
g(a)=\sqrt[3]{X^{3}}+\sqrt[3]{Y^{3}}=X+Y=\frac{ 1+\sqrt{x} }{2}
+\frac{ 1-\sqrt{x} }{2} =1.
\end{equation*}
Hence $\max_{a\geq 1/8}g(a)=1.$
