Matrices with $A^3+B^3=C^3$ Problem: Find infinitely many triples of nonzero $3\times 3$ matrices $(A,B,C)$ over the nonnegative integers with 
$$A^3+B^3=C^3.$$
My proposed solution is in the answers.

OR OR OR, given 
$$ x, y > 0,   $$
let
$$ R \; = \;  
 \left(  \begin{array}{ccc}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  x & 0  & 0  
\end{array} 
  \right)  , \; \;
 S \; = \;  
 \left(  \begin{array}{ccc}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  y & 0  & 0  
\end{array} 
  \right)  , \; \;
 T \; = \;  
 \left(  \begin{array}{ccc}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  x + y & 0  & 0  
\end{array} 
  \right)  ,
  $$
then
$$ R^3 = x I, \; \; S^3 = y I, \; \; T^3 = (x+y) I    $$
and
$$ R^3 + S^3 = T^3.  $$
OR
$$ S \; = \;  
 \left(  \begin{array}{rrr}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  2n^2 & 0  & 0  
\end{array} 
  \right)  , \; \;
 T \; = \;  
 \left(  \begin{array}{rrr}
  0 & 0 & 1 \\
  2 n  & 0 & 0 \\
  0 & 2 n  & 0  
\end{array} 
  \right) .
  $$
Then 
$$  S^3 = 2 n^2 I, \; \;  T^3 = 4 n^2 I,   $$
and
$$ S^3 + S^3 = T^3.  $$
OR OR, given a Pythagorean triple
$$ a^2 + b^2 = c^2,   $$
let
$$ R \; = \;  
 \left(  \begin{array}{rrr}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  a^2 & 0  & 0  
\end{array} 
  \right)  , \; \;
 S \; = \;  
 \left(  \begin{array}{rrr}
  0 & 1 & 0 \\
  0  & 0 & 1 \\
  b^2 & 0  & 0  
\end{array} 
  \right)  , \; \;
 T \; = \;  
 \left(  \begin{array}{rrr}
  0 & 0 & 1 \\
  c  & 0 & 0 \\
  0 & c  & 0  
\end{array} 
  \right) ,
  $$
then
$$ R^3 = a^2 I, \; \; S^3 = b^2 I, \; \; T^3 = c^2 I    $$
and
$$ R^3 + S^3 = T^3.  $$
