"Nearly" Fermat triples: case cubic Suppose $a^2+b^2-c^2=0$ are formed by a (integral) Pythagorean triple. Then, there are $3\times3$ integer matrices to generate infinitely many more triples. For example, take
$$\begin{bmatrix}-1&2&2 \\ -2&1&2 \\ -2&2&3\end{bmatrix}\cdot
\begin{bmatrix}a\\ b\\ c\end{bmatrix}=\begin{bmatrix}u\\ v\\ w\end{bmatrix}.$$
One may wish to do the same with $a^3+b^3-c^3=0$, but Fermat's Last Theorem forbids it!
Alas! one settles for less $a^3+b^3-c^3=\pm1$. Here, we're in good company: $9^3+10^3-12^3=1$, coming from Ramanujan's taxicab number $1729=9^3+10^3=12^3+1^3$. There are plenty more.

Question. Does there exist a concrete $3\times3$ integer matrix $M=[m_{ij}]$ such that whenever $a^3+b^3-c^3\in\{-1,1\}$ (integer tuple) then $u^3+v^3-w^3\in\{-1,1\}$ provided
  $$\begin{bmatrix}m_{11}&m_{12}&m_{13} \\ m_{21}&m_{22}&m_{23} \\ m_{31}&m_{32}&m_{33}\end{bmatrix}\cdot
\begin{bmatrix}a\\ b\\ c\end{bmatrix}=\begin{bmatrix}u\\ v\\ w\end{bmatrix}.$$

 A: Any such matrix $M$ would give rise to an automorphism of the cubic surfaces
$$a^3 + b^3 = c^3 \pm d^3  \quad \subset \mathbb{P}^3.$$
These are both just different ways of writing the Fermat cubic surface
$$x_0^3 + x_1^3 + x_2^3 + x_3^3 = 0 \quad \subset \mathbb{P}^3.$$
It is well-known that the automorphism group of the Fermat cubic surface over $\mathbb{C}$ is generated by the "obvious automorphisms", namely by permuting coordinates and multiplying coordinates by a third root of unity (See Table 9.6 of "Dolgachev - Classical algebraic geometry" for this claim).
Hence, as you are interested with matrices with integer entries, we see that the only such $M$ are the 6 permutation matrices which permute $a,b,c$, with also possibly multiplying some variable by $-1$ to fix the signs.
For example
\begin{bmatrix}0&1&0 \\ 1&0&0 \\ 0&0&1\end{bmatrix}
is an example of such a matrix. The other matrices can be written down analogously.
A: Daniel Loughran: I apologize for the confusion. You were right. I was only focusing on a single family of solutions to $x^3+y^3-z^3=\pm1$, for which of course there is tedious experimental finding that does a parametrization. Not for all solutions. 
Let's see if we can prove that it works: start with $[9,10,12]^T$ with $a^3+b^3-c^3=\pm1$ then I claim that $[u,v,w]^T=M^n\cdot[9,10,12]^T$ also satisfies $u^3+v^3-w^3=\pm1$.
Take the matrix to be
$$M=\begin{bmatrix}63&104&-68\\64&104&-67 \\80&131&-85\end{bmatrix}.$$
For example, if $[a,b,c]^T=[9,10,12]^T$ we have $9^3+10^3-12^3=1$ and
$$[u,v,w]^T=M\cdot[9,10,12]^T=[791,812,1010]^T$$ 
satisfies $791^3+812^3-1010^3=-1$. If we continue, 
$$M\cdot[791,812,1010]^T=[65601,67402,83802]^T$$
satisfies $65601^3+67402^3-83802^3=1$. This process generates and infinite family of solutions.
