Hilbert-Waring Theorem/Variant Form/Particular Cases/3

Particular Case of the Hilbert-Waring Theorem -- Variant Form: $k = 3$
The Hilbert-Waring Theorem -- Variant Form states that:

The case where $k = 3$ is:

Every sufficiently large positive integer can be expressed as the sum of a number of positive cubes.

The exact number is the subject of ongoing research, but at the time of writing ($11$th February $2017$) it is known that it is between $4$ and $7$.

That is:
 * $4 \le G \left({3}\right) \le 7$

Proof
From Cube Modulo 9, a cube is congruent to $0$, $1$, or $2$ modulo $9$.

So it takes at least $4$ cubes to make an integer $k \equiv \pm 4 \pmod 9$.

Hence $G \left({3}\right)$ is at least $4$.

For numbers less than $1.3 \times 10^9$, it has been demonstrated that $1 \, 290 \, 740$ is the highest number to require $6$ cubes.

It is also known that the number of numbers between $n$ and $2 n$ requiring $5$ cubes falls with increasing $n$. has expressed a belief that this decrease is rapid enough to suggest that $G \left({3}\right) = 4$.

From the work of, , and , $7 \, 373 \, 170 \, 279 \, 850$ is the largest number known that is not the sum of $4$ cubes. They offer an argument that this may indeed be the largest possible.