We have,
$$
\begin{aligned}
x & =\frac{2 \cdot 5}{3 \cdot 6}-\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{2}{5}\right)+\frac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\left(\frac{2}{5}\right)^2+\ldots \infty \\
& =\frac{2 \cdot 5}{3^2 \cdot 2 !}-\frac{2 \cdot 5 \cdot 8}{3^3 \cdot 3 !}\left(\frac{2}{5}\right)+\frac{2 \cdot 5 \cdot 8 \cdot 11}{3^4 \cdot 4 !}\left(\frac{2}{5}\right)^2
\end{aligned}
$$


multiply with $\left(\frac{2}{5}\right)^2$
$$
\begin{aligned}
& \frac{4}{25} x=\frac{2 \cdot 5}{3^2 \cdot 2 !}\left(\frac{2}{5}\right)^2-\frac{2 \cdot 5 \cdot 8}{3^3 \cdot 3 !}\left(\frac{2}{5}\right)^3+\ldots \infty \\
& \frac{4}{25} x-\frac{2}{3 \cdot 1 !}\left(\frac{2}{5}\right)+1=1-\frac{2}{3 \cdot 1 !}\left(\frac{2}{5}\right)+\frac{2 \cdot 5}{3^2 \cdot 2 !} \\
& \quad\left(\frac{2}{5}\right)^2-\frac{2 \cdot 5 \cdot 8}{3^3 \cdot 3 !}\left(\frac{2}{5}\right)^3+\ldots
\end{aligned}
$$
$$
\begin{aligned}
& \frac{4 x}{25}-\frac{4}{15}+1=\left(1+\frac{2}{5}\right)^{-\frac{2}{3}} \\
& \Rightarrow \quad \frac{4 x}{25}+\frac{11}{15}=\left(\frac{7}{5}\right)^{-\frac{2}{3}} \Rightarrow \frac{1}{5}\left(\frac{4 x}{5}+\frac{11}{3}\right)=\left(\frac{7}{5}\right)^{-\frac{2}{3}} \\
& \Rightarrow \frac{1}{5}\left(\frac{12 x+55}{15}\right)=\left(\frac{7}{5}\right)^{-\frac{2}{3}} \Rightarrow \quad \frac{12 x+55}{75}=\left(\frac{5}{7}\right)^{\frac{2}{3}}
\end{aligned}
$$
cube both sides,
$$
\begin{aligned}
\Rightarrow \frac{(12 x+55)^3}{(75)^3} & =\frac{5^2}{7^2} \quad \Rightarrow 7^2(12 x+55)^3=75^3 \cdot 5^2 \\
& =3^3 \cdot 5^6 \cdot 5^2=3^3 \cdot 5^8
\end{aligned}
$$