Diophantine Equation $a^6+b^6+c^6=d^6$ Within the literature on Diophantine equations there seems to be very little on the $6,1,3$ equation:
$$a^6+b^6+c^6=d^6\quad\quad(1)$$
Mathworld, for example, simply records that there are no known solutions of $6.1.n$ equations for $n \leq 6$. The former Euler Project searching for solutions of equations in equal sums of like powers showed $6,1,5$ at the top of its most wanted list but did not mention $6,1,3$.
One way of looking at the $6,1,3$ equation is as a special case of the much more familiar $3,1,3$ equation:
$$w^3+x^3+y^3=z^3\quad\quad(2)$$
in which each of the terms is also a square. It may be considered relevant that there are known solutions of $(2)$ in which two or three of the terms are square, the smallest respectively being:
$$(1^2)^3 + 6^3 + 8^3 = (3^2)^3\quad\quad(3)$$
$$118^3 + (15^2)^3 + (18^2)^3 = (19^2)^3\quad\quad(4)$$
Question: Are there any discussions in the literature of either:

*

*direct strategies for searching for solutions of $(1)$;

*research seeking a proof that $(1)$ has no non-trivial solutions?

By a direct strategy I mean one which addresses 6,1,3 itself, rather than one which searches for $6,1,n$ for higher $n$ with the remote possibility of finding a solution in which some of the terms are zero.  An example of a direct strategy would be to take the following parametric solution of $(2)$:
$$(9s^4)^3 + (3s(t^3-3s^3))^3 + (t(t^3-9s^3))^3 = (t^4)^3$$
which already has two square terms, and to try to find values of $s$ and $t$ such that the other two terms are also square.

COMMENT (This is not an answer!)
The identity $(2xz)^2+(2yz)^2+(z^2-x^2-y^2)^2=(x^2+y^2+z^2)^2$ gives the general solution of $X^2+Y^2+Z^2=W^2$ so one has the necessary condition with parameters $x,y,z$
$$\begin{cases}a^3=2xz\\b^3=2yz\\c^3=z^2-x^2-y^2\\d^3=x^2+y^2+z^2\end{cases}$$ This could be a good first step to get an answer.
(Follow the COMMENT).- By the way, if
$$A_1=a^5+10a^4-8a^3+16a^2+64a-32\\A_2=a^5-10a^4-8a^3-16a^2+64a+32\\A_3=-a^5+8a^4+8a^3-16a^2+80a+32\\A_4=-a^5-8a^4+8a^3+16a^2+80a-32$$ then we have for all real parameter $a$ the identity
$$a(6a^4+24a^2+96)^3=A_1^3+A_2^3+A_3^3+A_4^3$$ Consequently, making $a=n^3$, one has for all natural $n$ an identity:
$$X_1^3+X_2^3+X_3^3+X_4^3=X_5^3$$  where each $X_i$ has degree $15$.
