On $\eta(6z)\eta(18z)$ and the splitting / modularity of $x^3 - 2$ Consider one of the simplest non-abelian examples of modularity. Let $$\eta(6z)\eta(18z) = q\prod_{n=1}^\infty (1 - q^{6n})(1 - q^{18n}) = q - q^7 - q^{13} -q^{19} + q^{25} + 2q^{31} - q^{37} + 2q^{43} - q^{61} - q^{67} - q^{73} - q^{79} + q^{91} - q^{97} - q^{103} \dots = \sum_{n=1}^\infty a_n q^n$$
Then $a_p + 1 =$ the number of solutions of $\{x^3 - 2 = 0\}$ in $\mathbb{Z}/p\mathbb{Z}$, for prime $p > 3$.
1) What might be the best method to prove this? 
2) Is it possible to prove it without knowledge of modular forms? i.e. without using the dimension / basis of the relevant space of moduli forms of some weight and level and nebentypus. For example, can it be explained using Weyl–Kac character formula etc.?
3) Unfortunately LMFDB does not yet have a list of weight $1$ modular forms and their corresponding Artin representations. Where can we find more of such weight $1$ examples?
 A: Here's an answer to 2. You can tell me if it's the "best method".
Euler's Pentagonal number theorem gives that
$$ \eta(24z) = \sum_{n \in \mathbb{Z}} (-1)^{n} q^{(6n+1)^{2}}. $$
This yields the formula
$$ \eta(6z) \eta(18z) = \sum_{m, n \in \mathbb{Z}} (-1)^{m+n} q^{\frac{(6n+1)^{2} + 3(6m+1)^{2}}{4}}. $$
If $m=x$ and $n = x+2y$, then $m+n$ is even and we get
$$ \sum_{x, y \in \mathbb{Z}} q^{(6x+3y+1)^2 + 27y^{2}}. $$
On the other hand if $m=x$ and $n=x+2y+1$, then $m+n$ is odd and we get
$$ -\sum_{x,y \in \mathbb{Z}} q^{4(3x+y+1)^{2} + 2(3x+y+1)(2y+1) + 7(2y+1)^{2}}.$$
The upshot (skipping a few small details) is that
$$ \eta(6z) \eta(18z) = \frac{1}{2} \left[ \sum_{x,y \in \mathbb{Z}} q^{x^{2} + 27y^{2}} - q^{4x^{2} + 2xy + 7y^{2}} \right]. $$
Now, Gauss proved that if $p \equiv 1 \pmod{3}$ is prime, then there is some $z$ so that $z^{3} \equiv 2 \pmod{p}$ if and only if $p = x^{2} + 27y^{2}$ for some $x, y \in \mathbb{Z}$. It follows from this that the $p$th coefficient of $\eta(6z) \eta(18z)$ is $2$ if $p \equiv 1 \pmod{3}$ and $2$ is a cube modulo $p$. 
If $p \equiv 1 \pmod{3}$, but $2$ is not a cube modulo $p$, then $p$ is represented by $4x^{2} + 2xy + 7y^{2}$ in two ways, and this means that the $p$th coefficient of $\eta(6z) \eta(18z)$ is $-1$.
If $p \not\equiv 1 \pmod{3}$, the $p$th coefficient of $\eta(6z) \eta(18z)$ is clearly $0$. 
