Solve $\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$ by algebraic methods I was trying to solve this equation without using calculus.
Is it possible to be solved by elementary algebraic methods?
$$\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$$

Let $u=\sqrt[3]{q+\sqrt{p^{3}+q^{2}}}$, $v=\sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$,
we have $y=u+v$ satisfying $y^{3}+3py=2q \ldots \ldots (*)$.
Take $y=\sqrt[3]{2}$, $q=7x$ and $p^{3}+q^{2}=19^{2}$ or $p^{3}=19^{2}-(7x)^{2}$.
Substitute into $(*)$,
\begin{align*}
  2+3\sqrt[3]{19^{2}-(7x)^{2}} \times \sqrt[3]{2} &= 2(7x) \\
  \sqrt[3]{19^{2}-(7x)^{2}} &= \frac{14x-2}{3\sqrt[3]{2}} \\
  19^{2}-(7x)^{2} &= \frac{(14x-2)^{3}}{54}
\end{align*}
By factor theorem or Cardano formula,
$\displaystyle x=\frac{7}{4}, \frac{-8\pm 3i\sqrt{15}}{7}$ subject to correct choices of cubic roots in the original equation.
P.S.:  May rewrite as $\sqrt[3]{19+7x}-\sqrt[3]{19-7x}=\sqrt[3]{2}$ for easy computer verification.
