The proof relies on the method of infinite descent. If there is a triple, we can always find a smaller one, and that is a contradiction. The metric is the absolute value of the largest term. This is what gets "smaller".
Note that the coprime lemma is in force, hence we can assume the triples are coprime.
An earlier theorem shows 3 divides something. Move that term to the right and relable, so that z is divisible by 3. This rearrangement may make some of the variables negative, but that's all right.
Refer to the notation used in the previous section. In particular, you should be familiar with the variables q a b and c.
Our triple factors in the ring Z[q], and each factor is a cube, except for x+y, which is 9 times a cube. The latter is assured by the xyn lemma.
We're not sure which associate to use. That is, x+yq could be 1, q, or q2 times (u+vq)3×(1-q). Follow these possibilities through, and x+y is 3 times (a-b, b-c, or c-a). Again, I'm referring to the a b c notation in the previous section. Since x+y is 9 times a cube, we have a-b, b-c, or c-a is 3 times a cube. Let's substitute and see what happens, starting with a-b.
u3 + v3 - 3u2v = 3g3
Reduce mod 3, and u = -v. With a = u3+v3, a is divisible by 3, and b and c are also divisible by 3, making x and y divisible by 3. This is not a coprime solution.
The same problem occurs with u3+v3-3uv2, so move to the last case.
3u2v - 3uv2 = 3g3
uv(u-v) = g3
Since u and v are coprime, each is coprime to u-v. Thus all three are perfect cubes. The sum of two cubes, namely v and u-v, yields another cube, namely u. We only need show this is a smaller triple.
Remember that z is 3 times u2-uv+v2. Let u and v have opposite signs. Thus u-v is larger than u or v. Swap variables if necessary, so that u is positive. Now u(u-v) is larger than u-v, and adding v2 only makes this bigger. Yet z is bigger than that, so the implied triple is smaller.
Next assume u and v are positive. If u is larger, multiply by (u-v), a positive number, then add v2. This is still less than z. Similar reasoning holds if v is larger than u.
Finally let u and v be negative. Again, either u or v has the largest absolute value. Make them positive, and we have not changed their absolute values, nor have we changed u2-uv+v2. Thus |u| and |v| are still smaller than |z|.
Each triple leads to a smaller triple, and that is a contradiction. Two cubes cannot sum to a cube, and Fermat's last theorem holds for n = 3.
If two factors of 1-q divide x+yq then 3 divides x+yq, and x and y are divisible by 3. Thus there is one factor of 1-q, and x+yq is an associate of (u+vq)×(1-q).
Once again x+y is 3 times (a-b, b-c, or c-a). This has to be a perfect cube. In fact it has to be 27 times a perfect cube. Thus (a-b, b-c, or c-a) is 9 times a cube.
We showed above that a-b won't do, for if it is divisible by 3, then so are x and y. Similarly for c-a, hence we are left with b-c, or 3u2v-3uv2. This is 9 times a cube, hence uv(u-v) is 3 times a cube.
Since the factors are coprime, two are cubes and one is 3 times a cube. Thus two cubes sum to 3 times a cube; another triple. Is this one smaller?
Once again z is 3(u2-uv+v2). The reasoning given earlier on this page is valid here. The largest of u, v, and u-v is bounded by z, which is bounded by 3z3. The triples descend forever, and that is a contradiction. There is no solution.