= Solution
One form of the <Theorem of the Cube> is the cube principle: a <line bundle> $M$ on $E^3$ which is trivial on each of $E\times E\times\{O\}$, $E\times\{O\}\times E$, and $\{O\}\times E\times E$ is trivial. Here is a proof for a complex <elliptic curve>.
Since $E$ is a real two-dimensional torus, its integral cohomology has no torsion. The <Künneth theorem> decomposes
$$
H^2(E^3,\mathbb Z)=\bigoplus_{j=1}^3p_j^*H^2(E,\mathbb Z)\ \oplus\!\bigoplus_{i<j}H^1(E,\mathbb Z)\otimes H^1(E,\mathbb Z).
$$
Every single-factor summand is detected on a face containing that factor, and every cross summand is detected on the corresponding two-factor face. Restriction of $M$ to those faces is trivial, so its <First Chern class> has zero component in every summand. Thus $c_1(M)=0$.
The <holomorphic exponential sequence> identifies the kernel of $c_1$ in the <holomorphic Picard group> with $H^1(E^3,\mathcal O)/\operatorname{im}H^1(E^3,\mathbb Z)$. First cohomology decomposes factorwise here: $H^1(E^3,\mathcal O)=\bigoplus_jp_j^*H^1(E,\mathcal O)$ and likewise for integral first cohomology. Analytically the former decomposition says that the harmonic representatives of $(0,1)$-cohomology on the flat torus are the constant forms $d\overline z_j$, one from each factor. Consequently this kernel is $\operatorname{Pic}^0(E)^3$, and
$$
M\cong p_1^*M_1\otimes p_2^*M_2\otimes p_3^*M_3
$$
for degree-zero line bundles $M_j$ on $E$. Restrict to each coordinate axis. The hypothesis makes every $M_j$ trivial, so $M$ is trivial. This proves the <cohomological proof of the cube theorem for elliptic curves>; the holomorphic isomorphism is algebraic on these projective varieties by <GAGA>.
To state its usual addition-map consequence, write $m_I:E^3\to E$ for summation of the coordinates indexed by $I$ and $p_j=m_{\{j\}}$. For any line bundle $L$ on $E$,
$$
\boxed{m_{123}^*L\otimes p_1^*L\otimes p_2^*L\otimes p_3^*L
\cong m_{12}^*L\otimes m_{13}^*L\otimes m_{23}^*L.}
$$
More canonically, include the inverse of the constant bundle with fibre $L_O$ in the alternating product. Its restriction to a zero-coordinate face has pairwise cancellations, leaving the trivial bundle. The cube principle proves its triviality, which gives the displayed isomorphism after choosing a trivialization of the constant fibre. Pullback along any triple of maps to $E$ gives the corresponding alternating identity for their seven nonempty sums.
A pointwise divisor version can also be seen directly from Question 1. For a divisor $D$ and fixed points $a,b,c$, the alternating translation divisor
$$
T_{a+b+c}^*D-T_{a+b}^*D-T_{a+c}^*D-T_{b+c}^*D+T_a^*D+T_b^*D+T_c^*D-D
$$
has degree zero and point sum zero: translation by $t$ subtracts $\deg(D)t$ from its point sum, and all the alternating terms cancel. The principality criterion proves that this divisor is principal.
Over a <number field> $K$, use algebraic line bundles and maps defined over $K$, and use Galois-invariant divisors rather than arbitrary individual complex points. \b[The geometric line-bundle cube identity itself is still exact over $K$.] To see descent in this setting, first pass to an algebraic closure, or an embedding in $\mathbb C$. The alternating bundle becomes trivial there by the preceding proof. Choose its nowhere-zero section with value one in a chosen $K$-basis at $(O,O,O)$. Any Galois conjugate of that section differs by an invertible global function, hence a constant on the connected projective variety; its normalized value at the rational origin forces that constant to be one. The section therefore descends, giving a $K$-defined trivialization.
For the arithmetic formulation used with heights, the necessary modification is a uniformly bounded error. A chosen <Weil height machine> representative $h_L$ satisfies
$$
\boxed{h_L(P+Q+R)-h_L(P+Q)-h_L(P+R)-h_L(Q+R)+h_L(P)+h_L(Q)+h_L(R)-h_L(O)=O(1),}
$$
uniformly for $P,Q,R\in E(K)$. Indeed, pullback and tensor product of line bundles correspond to composition and addition of heights only modulo bounded functions. This is the <arithmetic cube identity for Weil heights>. Arbitrary naive heights do not satisfy an exact numerical cube identity; the <canonical height of an elliptic curve> removes the bounded error, as explained in Question 4. This arithmetic qualification does not weaken the exact geometric theorem.
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