A well known conjecture states that $$\{x^3+y^3+z^3:\ x,y,z\in\mathbb Z\}=\{m\in\mathbb Z:\ m\not\equiv\pm4\pmod 9\}.$$ For $m=33,\, 42$ an integer solution to the equation $x^3+y^3+z^3=m$ was only found last year.
In 2017, Tyrell asked whether $$\left\{\frac{x(x+1)(x+2)}6+\frac{y(y+1)(y+2)}6+\frac{z(z+1)(z+2)}6:\ x,y,z\in\mathbb Z\right\}=\mathbb Z,$$ see the question with the website http://math.stackexchange.com/questions/2472205. Few weeks ago Alkan (cf. Numbers of the form $x^2(x-1) + y^2(y-1) + z^2(z-1)$ with $x,y,z\in\mathbb Z$) conjectured that $$\left\{\frac{x^2(x-1)}2+\frac{y^2(y-1)}2+\frac{z^2(z-1)}2:\ x,y,z\in\mathbb Z\right\}=\mathbb Z.$$
I think it's interesting to find a cubic polynomial $P(x)$ with integer coefficients such that $$\{P(x)+P(y)+P(z):\ x,y,z\in\mathbb Z\}=\mathbb Z.$$ This led me to pose the following conjecture.
Conjecture. Each $m\in\mathbb Z$ can be written as a sum of three numbers of the form $x^3-2x\ (x\in\mathbb Z)$. In other words, we have $$\{x^3-2x+y^3-2y+z^3-2z: x,y,z\in\mathbb Z\}=\mathbb Z.$$
As $P(x)=x^3-2x$ is an odd function, the conjecture can be reduced to the case $m\in\mathbb N=\{0,1,2,\ldots\}$. Via computation I found that those natural numbers $n\le1000$ not in the set $$\{x^3-2x+y^3-2y+z^3-2z:\ x,y,z\in\{-1000,\ldots,1000\}\}$$ are \begin{gather}70,\ 75,\ 83,\ 86,\ 139,\ 185,\ 198,\ 237,\ 253,\ 262,\ 275, \ 305,\ 338,\ 355,\ 362, \\397, 414,\ 415,\ 422,\ 426,\ 457,\ 458,\ 509,\ 535,\ 558,\ 562,\ 564,\ 580,\ 583, \\ 593, \ 613,\ 614,\ 635,\ 642,\ 673,\ 677,\ 684, \ 693,\ 697,\ 722,\ 735,\ 779,\ 782, \\ 790,\ 791,\ 793,\ 807,\ 818,\ 850,\ 851,\ 870,\ 888,\ 898,\ 908,\ 943,\ 957. \end{gather} Let $S$ denote the set of these numbers.
QUESTION. Can we find an explicit solution of the equation $$n=x^3-2x+y^3-2y+z^3-2z\ \ (x,y,z\in\mathbb Z)$$ for each $n\in S$?
