An Ultimate Chinese High-School Maths Problem
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 require
\[
\begin{aligned}
\frac{8a-1}{3} &\geq 0\\
a &\geq \frac{1}{8}.
\end{aligned}
\]
Let
\[
\xi=
\sqrt[3]{a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}},
\]
and
\[
\zeta=
\sqrt[3]{a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}}.
\]
Then, we obtain
\[
\begin{aligned}
\xi\zeta
&=
\sqrt[3]{a^{2}-\frac{(a+1)^{2}}{9}\cdot\frac{8a-1}{3}}\\
&=
\sqrt[3]{\frac{27a^{2}-(a+1)^{2}(8a-1)}{27}}\\
&=
\sqrt[3]{\frac{-8a^{3}+12a^{2}-6a+1}{27}}\\
&=
\sqrt[3]{-\frac{(2a-1)^{3}}{27}}\\
&=
-\frac{1}{3}(2a-1).
\end{aligned}
\]
Hence,
\[
\begin{aligned}
(\xi+\zeta)^{3}
&=
\xi^{3}+\zeta^{3}+3\xi^{2}\zeta+3\xi\zeta^{2}\\
&=
2a+3\xi\zeta(\xi+\zeta)\\
&=
2a-(2a-1)(\xi+\zeta).
\end{aligned}
\]
Let \(\Gamma=\xi+\zeta\). Then, from above,
\[
\begin{aligned}
\Gamma^{3}+(2a-1)\Gamma-2a &= 0\\
\Gamma^{3}-\Gamma+2a\Gamma-2a &= 0\\
\Gamma(\Gamma^{2}-1)+2a(\Gamma-1) &= 0\\
\Gamma(\Gamma+1)(\Gamma-1)+2a(\Gamma-1) &= 0\\
(\Gamma-1)(\Gamma^{2}+\Gamma+2a) &= 0.
\end{aligned}
\]
Therefore,
\[
\begin{aligned}
\Gamma_{1} &= 1,\\
\Gamma_{2,3} &= \frac{-1\pm\sqrt{1-8a}}{2}.
\end{aligned}
\]
Considering \(\Gamma_{2,3}\), since we have already established that \(\displaystyle a\geq\frac{1}{8}\), for \(\Gamma_{2,3}\) to be real we must also have
\[
\begin{aligned}
1-8a &\geq 0\\
a &\leq \frac{1}{8}.
\end{aligned}
\]
Thus, both conditions can hold simultaneously only when
\[
a=\frac{1}{8}.
\]
At \(a=\frac{1}{8}\), the two roots from the quadratic factor are
\[
\Gamma_{2,3}
=
\frac{-1\pm\sqrt{1-8\cdot\frac18}}{2}
=
-\frac12.
\]
However, substituting \(a=\frac18\) directly into the original expression gives
\[
\begin{aligned}
\Gamma
&=
\sqrt[3]{\frac18+\frac{3/8}{3}\sqrt{0}}
+
\sqrt[3]{\frac18-\frac{3/8}{3}\sqrt{0}}\\
&=
\sqrt[3]{\frac18}
+
\sqrt[3]{\frac18}\\
&=
\frac12+\frac12\\
&=
1.
\end{aligned}
\]
Therefore, \(\Gamma_{2,3}=-\frac12\) do not correspond to the value of the original radical expression. The valid solution is \(\Gamma=1\).
To sum up,
\[
\Gamma=
\begin{cases}
\text{undefined over the reals}, & a<\frac18,\\
1, & a\geq\frac18.
\end{cases}
\]
Hence,
\[
\sqrt[3]{a+\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}}
+
\sqrt[3]{a-\frac{a+1}{3}\cdot\sqrt{\frac{8a-1}{3}}}
=
1,
\qquad
a\geq\frac18.
\]
