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(This summarizes the accepted answer of Somos, and uses just the j-function and Dedekind eta function.)

Given the j-function $j(\tau)$,

$$j(\tau) = 1728J(\tau)$$

implemented in Mathematica as j(t) = 1728KleinInvariantJ[t], then Klein's septic resolvent reduces to the elegant formula,

$$y\Big(y^3-\frac{8}{\alpha^3}\sqrt{-7}\Big)\Big(y^3-\sqrt{-7}\Big)=\sqrt[3]{j(\tau)}$$

where the seven roots are,

$$y_k = \frac{P_1(k)+\alpha P_2(k)}{\eta^2(\tau)}\\ \alpha=\frac{-1+\sqrt{-7}}2$$

for $k=1,2\dots7,$ with $P_1(k)$ and $P_2(k)$ as defined by Klein (and Somos).

(This summarizes the accepted answer of Somos, and uses the j-function only.)

Updated: Jan 16, 2023

I just realized that since Somos' seven roots $r_k$ has an eta function $\eta(\tau)$ as a denominator, this can be incorporated into Ramanujan's parameterization for $\lambda, \mu, \nu$ so they are theta quotients and clearly radicals for $\tau=\sqrt{-d}$. Re-define,

$$a = \frac{-q^{25/56}f(-q,-q^6)}{\quad\eta(\tau)} = \frac{-q^{17/42}f(-q,-q^6)}{\quad f(-q)}$$ $$b = \frac{q^{9/56}f(-q^2,-q^5)}{\quad\eta(\tau)} =\; \frac{q^{5/42}f(-q^2,-q^5)}{\quad f(-q)}$$ $$c = \frac{q^{1/56}f(-q^3,-q^4)}{\quad\eta(\tau)} = \frac{q^{-1/42}f(-q^3,-q^4)}{\quad f(-q)}$$

which also satisfy,

$$a^3b+b^3c+c^3a = 0$$

Then Klein's septic resolvent reduces to the elegant formula for the j-function $j(\tau)$,

$$z\Big(z^3-\frac{8}{h^3}\sqrt{-7}\Big)\Big(z^3-\sqrt{-7}\Big)=\sqrt[3]{j(\tau)}$$

where $h=\frac{-1+\sqrt{-7}}2$, and is solvable in radicals whenever $\tau = \sqrt{-d}.$ Its seven roots $z_k$ in terms of the theta quotients $a,b,c$ are,

$$z_k = R_1(k)+h\, R_2(k)$$ $$R_1(k) = \zeta^{k}\,a^2+\zeta^{4k}\,b^2+\zeta^{2k}\,c^2$$ $$R_2(k) = \zeta^{6k}\,ab+\zeta^{3k}\,bc+\zeta^{5k}\,ca$$

where $\zeta = e^{2\pi i/7}$ and for $k=1,2\dots7.$

(This summarizes the accepted answer of Somos, and uses just the j-function and Dedekind eta function.)

Given the j-function $j(\tau)$,

$$j(\tau) = 1728J(\tau)$$

implemented in Mathematica as j(t) = 1728KleinInvariantJ[t], then Klein's septic resolvent reduces to the elegant formula,

$$y\Big(y^3-\frac{8}{\alpha^3}\sqrt{-7}\Big)\Big(y^3-\sqrt{-7}\Big)=\sqrt[3]{j(\tau)}$$

where the seven roots are,

$$y_k = \frac{P_1(k)+\alpha P_2(k)}{\eta^2(\tau)}\\ \alpha=\frac{-1+\sqrt{-7}}2$$

with $P_1(k)$ and $P_2(k)$ for $k=1,2\dots7$ as defined by Klein (and Somos).

(This summarizes the accepted answer of Somos, and uses just the j-function and Dedekind eta function.)

Given the j-function $j(\tau)$,

$$j(\tau) = 1728J(\tau)$$

implemented in Mathematica as j(t) = 1728KleinInvariantJ[t], then Klein's septic resolvent reduces to the elegant formula,

$$y\Big(y^3-\frac{8}{\alpha^3}\sqrt{-7}\Big)\Big(y^3-\sqrt{-7}\Big)=\sqrt[3]{j(\tau)}$$

where the seven roots are,

$$y_k = \frac{P_1(k)+\alpha P_2(k)}{\eta^2(\tau)}\\ \alpha=\frac{-1+\sqrt{-7}}2$$

for $k=1,2\dots7,$ with $P_1(k)$ and $P_2(k)$ as defined by Klein (and Somos).

(This summarizes the accepted answer of Somos, and uses just the j-function and Dedekind eta function.)

Given the j-function $j(\tau)$,

$$j(\tau) = 1728J(\tau)$$

implemented in Mathematica as j(t) = 1728KleinInvariantJ[t], then Klein's septic resolvent reduces to the elegant formula,

$$y^7-7\left(\tfrac{-1+\sqrt{-7}}2\right)y^4-7\left(\tfrac{5+\sqrt{-7}}2\right)y-\sqrt[3]{j(\tau)}=0$$

where the seven roots are,

$$y_k = \frac{P_1(k)+\left(\frac{-1+\sqrt{-7}}2\right)P_2(k)}{\eta^2(\tau)}$$

with $P_1(k)$ and $P_2(k)$ for $k=1,2\dots7$ as defined by Klein (and Somos).

(This summarizes the accepted answer of Somos, and uses just the j-function and Dedekind eta function.)

Given the j-function $j(\tau)$,

$$j(\tau) = 1728J(\tau)$$

implemented in Mathematica as j(t) = 1728KleinInvariantJ[t], then Klein's septic resolvent reduces to the elegant formula,

$$y\Big(y^3-\frac{8}{\alpha^3}\sqrt{-7}\Big)\Big(y^3-\sqrt{-7}\Big)=\sqrt[3]{j(\tau)}$$

where the seven roots are,

$$y_k = \frac{P_1(k)+\alpha P_2(k)}{\eta^2(\tau)}\\ \alpha=\frac{-1+\sqrt{-7}}2$$

with $P_1(k)$ and $P_2(k)$ for $k=1,2\dots7$ as defined by Klein (and Somos).