The values of the parameter $\lambda_{ n }$ of Dedekind $\eta$-functions with $n$ even were calculated by Nipen Saikia in the article: "Some New Explicit Values of Parameters $\lambda_{ n}$ and $\mu$ of Quotients of Eta-Function", Hindawi P. C. ISRN Computational Mathematics, volume 2013, art. ID 435261, 8-pages".
$\lambda_{n}=\frac{1}{3\sqrt{3}}\frac{f(q)^{6}}{\sqrt{q}f(q^{3})^{6}}$
where
$q=e^{-\pi\sqrt{-\frac{n}{3}}}$.
and
$f(-q)=(1-q)(1-q^{2})(1-q^{3})(1-q^{4})(1-q^{5})...$
For example, the value of $\lambda_{ 2}$ is equal to:
$\lambda_{2}=\Big(-4+3\sqrt{3}+\sqrt{42-24\sqrt{3}}\Big)^{3/4}\Big(-44+27\sqrt{3}-3\sqrt{458-264\sqrt{3}}\Big)^{1/4}$
what else is but the following :
$\lambda_{2}=\Big(\sqrt{2}+1\Big)^{2}\Big(\sqrt{3}-\sqrt{2}\Big)^{3/2}$
It's much more elegant!
And so also
$\lambda_{ 4}=2^{-9/4}\Big(-76+33\sqrt{6}+\sqrt{12306-5016\sqrt{6}}\Big)^{3/4}\Big(\sqrt{-2+3\sqrt{6}}-\sqrt{-6+3\sqrt{6}}\Big)^{3/2}$
what else is but the following :
$\lambda_{ 4}=\Big(\sqrt{2}-1\Big)^{3}\Big(\frac{\sqrt{3}+1}{\sqrt{2}}\Big)^{9/2}$
It's much more elegant!
I feel compelled to give some values of $\lambda_{2n}$, calculated through Ramanujan's invariants $G_{n}$, and singular moduli of theory in signature 3.
$\lambda_{6}=3\sqrt{3}\big(\sqrt{2}-1\big)^{3/2}\big(2-\sqrt{3}\big)^{3/2}\big(\sqrt{3}+\sqrt{2}\big)^{2}$
$\lambda_{8}=\big(\sqrt{2}+1\big)^{5/2} \big(\sqrt{3}+\sqrt{2}\big)^{9/4} \Big(\sqrt{2+\sqrt{3}}-\sqrt{1+\sqrt{3}}\Big)^{3}$
$\lambda_{10}=\big(\sqrt{2}+1\big)^{3}\big(\frac{\sqrt{3}-1}{\sqrt{2}}\big(\frac{\sqrt{5+1}}{2}\big)^{6}\big(\sqrt{6}-\sqrt{5}\big)^{3/2}$
$\lambda_{14}=\big(5\sqrt{2}-7\big)\big(25\sqrt{7}-27\sqrt{6}\big)^{1/2}\big(9\sqrt{3}+11\sqrt{2}\big)\big(4\sqrt{14}+15\big)$ $\lambda_{16}=\big(1-\sqrt{2\sqrt{5}-4}\big)^{3}\big(\sqrt{2}+1\big)^{9/2}\big(\frac{\sqrt{3}+1}{\sqrt{2}}\big)^{21/4}\big(\sqrt{3}+\sqrt{2}\big)^{3}$
Who can help me find a simple expression for $\lambda_{12}$?
I think I found $\lambda_{12}=3^{3/4}\Big(\frac{\sqrt{3}+1}{\sqrt{2}}\Big)^{19/4}\Big(\sqrt{2}+3^{1/4}\Big)^{9/4}\Big(-3^{3/4}+\sqrt{3}-3^{1/4}+2\Big)^{3/2}\Big(\sqrt{3}-\sqrt{2}\Big)^{3/2}$.