Proving the sums of three cubes conjecture by the Hasse principle

Question

In his Cours d'arithmétique Serre applies the Hasse-Minkowski theorem to quadratic forms of the form:
$$
x^2 + y^2 + z^2 = n
$$
for $n \in \mathbb{N}$ to prove that a natural number $n$ is a square if and only if $-n$ is not a square in $\mathbb{Q}_2$, and hence that a natural number can be expressed as the sum of three squares provided that it is not of the form $4^a(8b-1)$ for some $a\in \mathbb{N}_0$, $b \in \mathbb{N}$.

My question is about if one might prove the sums of three cubes conjecture by analogous means.

Now we know that the Hasse principle does not hold for a general cubic forms, if we knew that it holds for cubic forms of the form:
$$
x^3 + y^3 + z^3 = n
$$
for some $n \in \mathbb{N}$, could we reduce the proof of the sum of three cubes conjecture to a local statement, analogous to the above.

Has any work been done to prove the sum of the cubes conjecture by these means?

Is the Hasse principle known to hold for cubic forms of the above form?

I suppose this would naturally entail the introduction of something like an analogy of the Hilbert symbol in the cubic case.

Many thanks.

Edit: This would naturally not work if the Hasse principle were known to fail for a cubic form:
$$
x^3 + y^3 + z^3 = n
$$
for some $n \in \mathbb{N}$, so a natural way of dismissing the above proof strategy would be to provide such a counter-example. – Is any such counter-example known to exist?

Answer 1

I don't think that the Hasse principle is useful here. First, Hasse-Minkowski fails in general for cubic polynomials, e.g., for $3x^3+4y^3+5z^3$. For $x^3+y^3+z^3=n$ I don't see how to use it.

Secondly, Ramanujan has used generating functions to obtain parametrised solutions for $x^3+y^3+z^3=1$ and $x^3+y^3+z^3=2$, and this direction appears to be more interesting, see for example the article by Harper.

In $1992$, Roger Heath-Brown conjectured that every $n$ unequal to $4$ or $5$ modulo $9$ has infinitely many representations as sums of three cubes.Conversely it easy to see that the other $n$ cannot be a sum of three cubes. There are many computational aspects, too, which made the problem popular for $n=33$ and $n=42$.