It is rather well-known that,
$e^{\pi\sqrt{43}} \approx 960^3 + 743.999\ldots$
$e^{\pi\sqrt{67}} \approx 5280^3 + 743.99999\ldots$
$e^{\pi\sqrt{163}} \approx 640320^3 + 743.999999999999\ldots$
Not so well-known is,
$e^{\pi\sqrt{43}} \approx (5x_1)^3 + 6.000000010\ldots$
$e^{\pi\sqrt{67}} \approx (5x_2)^3 + 6.000000000061\ldots$
$e^{\pi\sqrt{163}} \approx (5x_3)^3 + 6.000000000000000034\ldots$
where the $x_i$ is the appropriate root of the sextics,
$5x^6-960x^5-10x^3+1 = 0$
$5x^6-5280x^5-10x^3+1 = 0$
$5x^6-640320x^5-10x^3+1 = 0$
One can see the j-invariants (or at least their cube roots) appearing again. These sextics are solvable in radicals, factoring over $Q(\sqrt{5})$. However, a more interesting field is $Q(\phi)$, with the golden ratio $\phi = (1+\sqrt{5})/2$. Hence, these sextics have the relevant cubic factor,
$5x^3 – 5(53+86\phi)x^2 + 5(\color{blue}{8}+\color{blue}{13}\phi)x – (\color{red}{18}+\color{red}{29}\phi) = 0$
$5x^3 – 20(73+118\phi)x^2 – 20(\color{blue}{21}+\color{blue}{34}\phi)x – (\color{red}{47}+\color{red}{76}\phi) = 0$
$5x^3 – 20(8849+14318\phi)x^2 + 20(\color{blue}{377}+\color{blue}{610}\phi)x – (\color{red}{843}+\color{red}{1364}\phi) = 0$
respectively. Compare the x term with the Fibonacci numbers,
$F_n = 0, 1, 1, 2, 3, 5, \color{blue}{8, 13, 21, 34}, 55, 89, 144, 233, \color{blue}{377, 610},\dots$
and the constant term with the Lucas numbers,
$L_n = 2, 1, 3, 4, 7, 11, \color{red}{18, 29, 47, 76}, 123, 199, 322, 521, \color{red}{843, 1364},\dots$
Why, oh, why?
P.S. These can be easily verified in Mathematica using the Resultant[] function,
Resultant[$5x^3 – 5(53+86\phi)x^2 + 5(8+13\phi)x – (18+29\phi)$, $\phi^2-\phi-1$, $\phi$]
which eliminates $\phi$ and restores the original sextic. (Similarly for the other two.)
Best Answer
Solve $5x^6-Ax^5-10x^3+1$ for A, and you get $A=5x-\frac{10}{x^2}+\frac{1}{x^5}$
Solve $A^3+744 = (5x)^3+6$ for A, and you get $A=(125x^3-738)^{\frac13}$
These are almost equal as $x\rightarrow\infty$. In fact, if the difference is $D(x)$, then taking the taylor series of $D(\frac1y)$ around $y=0$ gives $D(x)=-\frac{4}{25x^2}+\frac{63641}{3125y^5}+\dots$
This solves half your mystery.