Solve \[\sqrt[3]{a+\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}}\]


From the question, we regulate
\frac{8a-1}{3}&\geq 0\
a&\geq \frac{1}{8}

Let \[\xi = \sqrt[3]{a+\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}}\]
\[\zeta=\sqrt[3]{a-\frac{a+1}{3}\cdot \sqrt{\frac{8a-1}{3}}}\]

Then, we obtain
\xi \cdot \zeta &= \sqrt[3]{a^{2}-\frac{\left (a+1 \right )^{2}}{9}\cdot \frac{8a-1}{3}}\
&= \sqrt[3]{\frac{27a^{2} - \left (a+1 \right )^{2} \cdot \left (8a-1 \right )}{27}}\
&= \sqrt[3]{\frac{-8a^{3}+12a^{2}-6a+1}{27}}\
&= \sqrt[3]{\frac{-\left (2a-1 \right )^{3}}{27}}\
&= -\frac{1}{3}\left ( 2a-1 \right )
\end{align*} \]

\left (\xi+\zeta \right )^{3} &= \xi^{3}+\zeta^{3}+3\xi^{2}\zeta+3\xi\zeta^{2}\
&=2a+3\xi\zeta\left ( \xi+\zeta \right )\
&=2a-\left ( 2a-1 \right )\left ( \xi+\zeta \right )

Let \(\Gamma =\xi+\zeta\), then from above
\Gamma^{3}+\left ( 2a-1 \right )\Gamma-2a&=0\
\Gamma\left ( \Gamma^{2} -1\right )+2a\left (\Gamma-1 \right )&=0\
\Gamma\left ( \Gamma +1\right )\left ( \Gamma -1\right )+2a\left (\Gamma-1 \right )&=0\
\left ( \Gamma -1\right )\left ( \Gamma^{2}+\Gamma+2a \right )&=0

We get

\Gamma_{2,3}&=\frac{-1\pm \sqrt{1-8a}}{2}\

Considering \(\Gamma_{2,3}\), since we've regulated that \(a\geq \frac{1}{8}\), here however we have to confine that
1-8a&\geq 0\
a&\leq \frac{1}{8}
Thus in this specific case, we can only take \(a= \frac{1}{8}\),

To sum up,
\Gamma &= \left\{\begin{matrix}
\text{invalid}, & a < \frac{1}{8}\
\Gamma_{1}=1, & a \geq \frac{1}{8}\\
\Gamma_{2,3}=1, & a=\frac{1}{8}
&= \left\{\begin{matrix}
\text{invalid}, & a < \frac{1}{8}\
\Gamma_{1,2,3}=1, & a \geq \frac{1}{8}\

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