$$\begin{align}
\left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right) + 1 &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)f_6^2 \\
\left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right) - 1 &= \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)f_6^2
\end{align}$$ Or more simply $$a+1=b \qquad \\ a-1=c \qquad $$ From the above, we find \((a,b,c)\). Define,$$\begin{align}\frac{ab}c &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)^2 \\ \frac{bc}a &= \,\big(f_6\big)^4 \\ \frac{ac}b &= \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)^2 \end{align}$$ Then we have the analogous relation,
\left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right) + 1 &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)f_6^2 \\
\left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right) - 1 &= \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)f_6^2
\end{align}$$ Or more simply $$a+1=b \qquad \\ a-1=c \qquad $$ From the above, we find \((a,b,c)\). Define,$$\begin{align}\frac{ab}c &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)^2 \\ \frac{bc}a &= \,\big(f_6\big)^4 \\ \frac{ac}b &= \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)^2 \end{align}$$ Then we have the analogous relation,
$$\left(\frac{4a}{bc}+\frac{bc}a\right)+2a = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$
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