well the first thing to check is if x^{3}+ 2 is irreducible over Z_{7}. since it is a cubic, we need only check to see if it has a root in Z_{7}. working (mod 7) we have:

0^{3}+ 2 = 2

1^{3}+ 2 = 3

2^{3}+ 2 = 1 + 2 = 3

3^{3}+ 2 = 6 + 2 = 1

4^{3}+ 2 = (-3)^{3}+ 2 = -6 + 2 = -4 = 3

5^{3}+ 2 = (-2)^{3}+ 2 = -1 + 2 = 1

6^{3}+ 2 = (-1)^{3}+ 2 = -1 + 2 = 1

this tells us that if we adjoin any root of x^{3}+ 2 to Z_{7}, say u, then Z_{7}(u) is isomorphic to the field Z_{7}[x]/<x^{3}+ 2> (where u corresponds to the coset of x in the quotient ring).

this alone is no guarantee that x^{3}+ 2 splits in Z_{7}(u). however, if we find a second root in Z_{7}(u), say v, we must have the splitting field, since we can then divide x^{3}+ 2 by (x - u)(x - v) to get the third factor.

now considering Z_{7}(u) as a vector space over Z_{7}, we see that {1,u,u^{2}} is a basis (because of the irreducibility of x^{3}+ 2) so we know that Z_{7}(u) has 343 elements. we could (if we were masochistic) try the 335 elements that might be roots, but there is a better way.

as we saw above, 2^{3}= 1 (mod 7), so (2u)^{3}= 2^{3}u^{3}= u^{3}. this means that 2u is also a root of x^{3}+ 2. similarly, 4^{3}= (4^{2})(4) = (2)(4) = 1 (mod 7), so 4u is also a root of x^{3}+ 2, hence x^{3}+ 2 splits in Z_{7}(u) as:

x^{3}+ 2 = (x - u)(x - 2u)(x - 4u).

checking, we have that:

(x - u)(x - 2u) = x^{2}- 3ux + 2u^{2}= x^{2}+ 4ux + 2u^{2}, and:

(x^{2}+ 4ux + 2u^{2})(x - 4u) = (x^{2}+ 4ux + 2u^{2})(x + 3u)

= x^{3}+ 4ux^{2}+ 2u^{2}x + 3ux^{2}+ 5u^{2}x + 6u^{3}

= x^{3}+ 6u^{3}.

but since u^{3}= -2 = 5, 6u^{3}= (6)(5) = 2.

(this happy state of affairs occurs because Z_{7}happens to contain all 3 cube roots of 1).