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    $\begingroup$ Let $n = 1, 2$ and we have solutions for $x, y, z$ with a value $t \in \mathbb{N}$ however not all $x, y, z$ are positive, though I think this may serve as a little bit of help: $$\begin{align} 1 &= (9t^3 + 1)^3 + (9t^4)^3 + (-9t^4 - 3t)^3 \\ 2 &= (6t^3 + 1)^3 + (-6t^3 - 1)^3 + (-6t^2)^3 \end{align}$$ or for big solutions for $n = 1$: $$1 = (1 - 9t^3 + 648t^6 + 3888t^9)^3 + (-135t^4 + 3888t^{10})^3 + (3t - 81t^4 - 1296t^7 - 3888t^{10})^3$$ $\endgroup$ Commented Oct 27, 2017 at 5:38
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    $\begingroup$ The expression for n = 2 should read: $(6t^3 + 1)^3 + (- 6t^3 + 1)^3 + (- 6t^2)^3 = 2$ That is, the 2nd sign in the 2nd pair of parentheses should be + rather than -. $\endgroup$ Commented Nov 26, 2017 at 5:37
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    $\begingroup$ @joro: Using Huisman's results, one finds that $$x^3+y^3+z^3 = 972$$ already has $96$ solutions in the "small" finite range that he searched. Similarly for other $n$. It is tempting to speculate that, if the range were infinite, then it and others in fact have infinitely many solutions like $n=1$. $\endgroup$ Commented Dec 30, 2017 at 3:17
  • $\begingroup$ @TitoPiezasIII What if in general the solutions are exponentially growing? $\endgroup$ Commented Dec 30, 2017 at 9:15
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    $\begingroup$ The Wikipedia article Sums of three cubes contains some results on this and also quite long list of references. $\endgroup$ Commented Sep 18, 2019 at 5:42