Does $a^6+b^6 = c^6+d^3$ have a non-trivial solution? It is conjectured that,
$$x_1^8+x_2^8+x_3^8 = y_1^8+y_2^8+y_3^8\tag{1}$$
has no non-trivial solutions. However, if we relax it a bit then,
$$x_1^8+x_2^8+x_3^4 = y_1^8+y_2^8+y_3^4\tag{2}$$
can be shown to have an infinite number of primitive solutions such as,
$$1118^8 + 1937^8 + 2502045^4 = 455^8 + 1846^8 + 3200691^4$$
Likewise, it is conjectured that,
$$x_1^6+x_2^6 = y_1^6+y_2^6\tag{3}$$
is only trivially solvable. However,
$$x_1^6+x_2^3 = y_1^6+y_2^3\tag{4}$$
also has an infinite number of solutions, such as,
$$5^6+167^3 = 8^6+164^3$$
Question: How far can we relax $(3)$? Is
$$a^6+b^6 = c^6+d^3\tag{5}$$
non-trivially solvable? 
 A: Mea culpa. I should have done a Mathematica search before posting this question. Normally I would, but I thought if there was a non-trivial solution to, 
$$a^6+b^6 = c^6+d^3\tag{1}$$
then it might be large. Turns out it is small,
$$15^6+18^6 = 19^6 + (-118)^3$$
and is the only primitive one with $a,b,c,< 100$. (Beyond that, it takes too long for my rather slow computer.)
I'll have to revise my question to: Is there a solution $a,b,c,d$ all positive?
Edit (a day later)
I just realized, regardless of the positivity of $a,b,c,d$, that there is an infinite number of primitive solutions to $(1)$ using the well-known identity,
$$(3t^2)^6+(3t-9t^4)^3 = 1+(9t^3-1)^3$$
All one has to do is find rational t such that the second term is also a square, 
$$3t-9t^4 = y^2$$
which, after transforming to Weierstrass form, is an elliptic curve. One such point is $t=3/13$ which yields,
$$27^6+138^6 = 13^{12}+(-25402)^3$$
Another is $t = 208098704151/634343459737$ (though presumably there may be points of smaller height), and so on.
