Apart from $(3, 4, 5, 6)$ are there any more primitive solutions to $x^3+y^3+z^3=w^3$ where $x^2+y^2=z^2$ ?
I’ve noted that if $\gcd(x ,y ,z) = k$, then $k$ divides $w$, so non-primitive Pythagorean triples are irrelevant.
The set of primitive Pythagorean triples is given by, $$\pm x= a^2-b^2$$ $$\pm y=2ab$$ $$\pm z= a^2+b^2$$ where $gcd(a,b)=1$ and $a, b$ are of opposite parity (one odd, one even).
So w is given by one of the following $$w^3=2a^2(a^4+4ab^3+3b^4) $$ $$w^3=2a^2(-a^4+4ab^3-3b^4) $$ $$w^3=2a^2(a^4-4ab^3+3b^4) $$ $$w^3=2b^2(3a^4+4ab^3+b^4) $$ $$w^3=2b^2(-3a^4+4ab^3-b^4) $$ $$w^3=2b^2(3a^4-4ab^3+b^4) $$
I’ve written a program to generate $(x,y,z)$ from $(a,b)$ with $b < a$ and of opposite parity, and then run it, with all the sign options, to $a = 41282$.
Edit 17 July 2016
If nobody objects, I should now like to restrict this question to positive integer values of $x, y, z$. I consider my original idea of allowing signed integers resulted in too many sub-problems, thus discouraging further progress.
Hence, the equation to be solved is $$w^3=2a^2(a^4+4ab^3+3b^4)$$
After the substitutions $a=2c$ and $w=2v$ this is
$$v^3= c^2 (b+2c)^2 (3b^2-4bc+4c^2)$$
with ($c$ odd, or even with the highest power of $2$ that divides $c$ of the form $3n$)
$b$ odd and $b<2c$.