Diophantin equation $a^3+b^3=c^3+5$

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When trying to solve another equation, I came up with this equation:

$$a^3 + b^3 = c^3 + 5, \space\space (a,b,c)\in\mathbb{Z}^3$$

It seems that it doesn't have any solutions. I tried to prove this. First I noticed that one of the variables must be even and then the other two must be of different parity. I plugged the according forms in but it didn't really lead me anywhere. I arrived at a equation:

$$2(k^3+n^3+8s^3) + 3s(4s+1) = 1$$

Is there any general theory for these kind of "Fermat's Last Theorem -equation with added constant"? Or how to solve this particular equation?

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It's quite trivial, actually. Every cube is either a multiple of 9, or one more or one less than a multiple of 9, as you see if you calculate $x^3$ and $(x ± 1)^3$. The sum or difference of three cubes is at most 3 away from a multiple of 9. 5 is four away from the nearest multiple of 9 and therefore not the sum or difference of 3 cubes.