Question

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


Solution:

From the question, we regulate
\[\begin{align*} \frac{8a-1}{3}&\geq 0\\ a&\geq \frac{1}{8}
\end{align*}\]

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

Hence,
\[\begin{align*} \left (\xi+\zeta \right )^{3} &= \xi^{3}+\zeta^{3}+3\xi^{2}\zeta+3\xi\zeta^{2}\\ &=2a+3\xi\zeta\left ( \xi+\zeta \right )\\ &=2a-\left ( 2a-1 \right )\left ( \xi+\zeta \right ) \end{align*}\]

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

We get

\[\begin{align*} \Gamma_{1}&=1\\ \Gamma_{2,3}&=\frac{-1\pm \sqrt{1-8a}}{2}\\ \end{align*}\]

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

To sum up,
\[\begin{align*} \Gamma &= \left\{\begin{matrix} \text{invalid}, & a < \frac{1}{8}\\ \Gamma_{1}=1, & a \geq \frac{1}{8}\\ \Gamma_{2,3}=1, & a=\frac{1}{8} \end{matrix}\right.\\ &= \left\{\begin{matrix} \text{invalid}, & a < \frac{1}{8}\\ \Gamma_{1,2,3}=1, & a \geq \frac{1}{8}\\ \end{matrix}\right.\\ \end{align*}\]