<s>[INST] On Shanks' quartic approximation $\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)$ In Mathworld's "Pi Approximations", (line 58), Weisstein mentions one by the mathematician Daniel Shanks that differs by a mere $10^{-82}$,
$$\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)\color{blue}{+10^{-82}}\tag{1}$$
and says that $u$ is a product of four simple quartic units, but does not explicitly give the expressions. I managed to locate Shanks' Dihedral Quartic Approximations and Series for Pi (1982) Quartic Approximations for Pi (1980) online before so,
$$u = (a+\sqrt{a^2-1})^2(b+\sqrt{b^2-1})^2(c+\sqrt{c^2-1})(d+\sqrt{d^2-1}) \approx 1.43\times10^{13}$$
where,
$$\begin{aligned}
a &= \tfrac{1}{2}(23+4\sqrt{34})\\
b &= \tfrac{1}{2}(19\sqrt{2}+7\sqrt{17})\\
c &= (429+304\sqrt{2})\\
d &= \tfrac{1}{2}(627+442\sqrt{2})
\end{aligned}$$
(Remark: In Shank's paper, the expressions for $a,b$ are different since he didn't express them as squares.)  
A small tweak to $(1)$ can vastly increase its accuracy to $10^{-161}$,
$$\pi \approx \frac{1}{\sqrt{3502}}\ln\big((2u)^6+24\big)\color{blue}{+10^{-161}}\tag{2}$$
I noticed the constant $u$ can also be expressed in terms of the Dedekind eta function $\eta(\tau)$ as,
$$u = \frac{1}{2}\left(\frac{\eta(\tfrac{1}{2}\sqrt{-3502})}{\eta(\sqrt{-3502})}\right)^4\approx 1.43\times 10^{13}$$
which explains why $24$ improves the accuracy. Note that the class number of $d = 4\cdot3502$ is $h(-d)=16$, and $u$ is an algebraic number of deg $16$. Mathworld has a list of class numbers of d. However, we can also use those with $h(-d)=8$ such as,
$$x = \frac{1}{\sqrt{2}}\left(\frac{\eta(\tfrac{1}{2}\sqrt{-742})}{\eta(\sqrt{-742})}\right)^2 \approx 884.2653\dots$$
which is a root of the 8th deg,
$$1 + 886 x + 1535 x^2 + 962 x^3 + 1628 x^4 - 962 x^5 + 1535 x^6 - 886 x^7 + x^8 = 0$$
Question: Analogous to $u$, how do we express $x$ as a product of two quartic units?
 [/INST] Let $a = \displaystyle{ \frac{11 + \sqrt{106}}{2}}$ and $b = \displaystyle{ \frac{21 + 2 \sqrt{106}}{2}}.$ Then
$$x = (a + \sqrt{a^2 - 1}) (b + \sqrt{b^2 + 1}).$$
As requested, this exhibits $x$ as a product of two quartic units. (For the purists, note that $a$ and $b$ are only half algebraic integers, but the expressions above are genuinely units. I wrote them in this form to conform with the previous example of the OP) The first unit (involving $a$) generates a degree four extension whose Galois closure has  Galois group $D_8$, but the latter generates a Galois extension with Galois group $\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$.
For those playing at home, this allows for the extra simplification
$$b + \sqrt{b^2 + 1} =  (1 + \sqrt{2})^2 \cdot \frac{7 + \sqrt{53}}{2} .$$
</s>