Let $r$ be the positive real solution to $x^3 + \frac{2}{5} x - 1 = x^3 - 3x + 7.$ Find the exact numerical value of
\[r^2 + 2r^5 + 3r^8 + 4r^{11} + \dotsb.\]
By symmetry, we can write
\begin{align*}
r^2 + 2r^5 + 3r^8 + 4r^{11} + \dotsb &= \frac{1}{1 + r^3} + 2 \cdot \frac{r^3}{1 + r^3} + 3 \cdot \frac{r^6}{1 + r^3} + 4 \cdot \frac{r^9}{1 + r^3} + \dotsb \\
&= \frac{1 + 2r^3 + 3r^6 + 4r^9 + \dotsb}{1 + r^3} \\
&= \frac{d}{dr} \left( \frac{1 + r^3 + r^6 + r^9 + \dotsb}{1 + r^3} \right).
\end{align*}Then
\begin{align*}
\frac{1 + r^3 + r^6 + r^9 + \dotsb}{1 + r^3} &= 1 + \frac{r^3}{1 + r^3} + \frac{r^6}{1 + r^3} + \frac{r^9}{1 + r^3} + \dotsb \\
&= 1 + \frac{r^3}{1 + r^3} + \frac{r^6}{1 + r^3} + \frac{r^9}{1 + r^3} + \dotsb \\
&= 1 + r^3 \left( \frac{1}{1 + r^3} + \frac{r^3}{1 + r^3} + \frac{r^6}{1 + r^3} + \dotsb \right) \\
&= 1 + \frac{r^3}{1 - r^6}.
\end{align*}Therefore,
\[\frac{d}{dr} \left( \frac{1 + r^3 + r^6 + r^9 + \dotsb}{1 + r^3} \right) = \frac{d}{dr} \left( 1 + \frac{r^3}{1 - r^6} \right) = \frac{d}{dr} \left( \frac{1 - r^6 + r^3}{1 - r^6} \right).\]Recall that we want to find the value of this expression when $r = \sqrt[3]{2} - 1.$ Then when $r = \sqrt[3]{2} - 1,$
\begin{align*}
\frac{1 - r^6 + r^3}{1 - r^6} &= \frac{1 - (\sqrt[3]{2} - 1)^6 + (\sqrt[3]{2} - 1)^3}{1 - (\sqrt[3]{2} - 1)^6} \\
&= \frac{1 - (2 \sqrt[3]{2} - 6 \sqrt[3]{2} + 6) + (2 \sqrt[3]{2} - 3)}{1 - (2 \sqrt[3]{2} - 6 \sqrt[3]{2} + 6)} \\
&= -\frac{(\sqrt[3]{2} - 7)(\sqrt[3]{2} - 8)}{(\sqrt[3]{2} - 3)(\sqrt[3]{2} - 2)} \\
&= \boxed{8}.
\end{align*}