To recall, there are infinitely many primitive solutions to, $$a^4+b^8+c^8 = d^4+e^8+f^8$$ such as, $$14805^4 + 86^8 + 149^8 = 18939^4 + 35^8 + 142^8$$
discussed here. This almost violates the Lander-Parkin conjecture (which states that for ESLP with $k>3$, at least $k$ terms are needed).
I. Sixth Powers
Going higher, it turns out there are infinitely many primitive solutions to,
$$x_1^6+x_2^6+x_3^{\color{blue}{12}} = y_1^6+y_2^6+y_3^{\color{blue}{12}}$$
small ones of which are,
$$\; 179^6 + 275^6\, + 6^{\color{blue}{12}} = 276^6 + 65^6 + 13^{\color{blue}{12}}$$ $$111^6 + 230 ^6 + 11^{\color{blue}{12}} = 26^6 + 13^{\color{blue}{12}} + 15^{\color{blue}{12}}$$
To find infinitely more of the first kind, we use the $6$th-power parameterization with the smallest known degree, the Brudno-Delorme Identity,
\begin{align} a &= -n^4 - n^3 - 5n^2 + 8n + 8\\ b &= (n^3 + 7n - 2)(n + 2)\\ c &= 9n^2 + 6n + 12\\ d &= -n^4 + n^2 + 14n + 4\\ e &= -4n^3 - 5n^2 - 8n + 8\\ f &= (n^2 - n + 3)(n + 2)^2 \end{align}
which obeys at least three relations,
$$a^6+b^6+c^6 = d^6+e^6+f^6$$ $$a^2+b^2+c^2 = d^2+e^2+f^2$$ $$3a+b+c = 3d+e+f$$
To make two of its terms as squares, we simply solve,
\begin{align} 9n^2 + 6n + 12 &= \square_1\\[4pt] n^2 - n + 3 &= \square_2\end{align}
starting with initial rational point $n = -13/24.$ Converting this into an elliptic curve, one can then find infinitely more points.
II. Question
Since,
$$a^6 + b^6 + c^{\color{blue}{12}} = d^6 + e^{\color{blue}{12}} + f^{\color{blue}{12}}$$
has at least one non-trivial solution given above, can we in fact also solve,
$$a^6 + b^{\color{blue}{12}} + c^{\color{blue}{12}} = d^6 + e^{\color{blue}{12}} + f^{\color{blue}{12}}$$
where $(a,b,c)\neq(d,e,f)$?
