The standard way to answer questions like this is to use "modular arithmetic". The idea is that every integer is equal to
6n+r where r=0,1,2,3,4, or 5; and that addition, subtraction, and multiplication work on these remainders in the obvious way. For instance, if a = 6n+3 and b = 6m+2, a+b = 6p + 5. If the remainder is too big just take away sixes till it fits: (6n + 5) + (6m + 4) = 6p + 3 (since 9 = 6+3).
Because 6 is not prime, division does not work! It does for prime "bases".

There are shorter notations that we can use here. We can say (eg) 26 is "equivalent to 2 modulo 6", or we can say it belongs to the equivalence class [2].

Now work case by case. For instance,

[2]³ = [2][4] = [2] (the class containing 8)
11*[2] = [5][2] = [4] (the class containing 10)
[2]³ + 11*[2] = [0] (the class containing 6, and all
the numbers divisible by 6.

Do the same for [0],[1],[3],[4],[5] and you're done.

It's hardly worth doing here, but for numbers that aren't prime powers you can factor and check each individually: in other words, to show x³ + 11x is divisible by 6, show it's divisible by 2 and by 3.

Good Hunting!
RD