Taking $\pmod{7}$ we get $x^3\equiv 2u^3$, which implies $7|x,u$ [the only cubic residues are $0$ and $\pm 1$], or in other words $x,u=7x',u'$. Dividing everything in the equation by $7$ we get back to the original equation but with smaller sum of absolute values, which by infinite descent* implies all number were $0$, so that $(0,0,0,0,0,0)$ is the only solution. *If $x$ and $u$ were both already $0$, dividing by 7$ would give us a very similar equation which can be solved in a very similar way

Well, first of all you mixed up the names of the variables, I think. Secondly and slightly more severely (but still mostly harmless), your claim is not quite true because it could be that the two variables you shrink were zero (but not everyone is zero) and so the sum of absolute values does not decrease. So what you should do to receive a full score is either to just repeat the divisibility argument three times or to find a slightly more sophisticated semi-invariant that actually does decrease for sure.

Tintarn wrote: Determine all $6$-tuples $(x,y,z,u,v,w)$ of integers satisfying the equation \[x^3+7y^3+49z^3=2u^3+14v^3+98w^3.\] $\pmod{7}$ gives $7\mid x,u$. Then $\pmod{7^2}$ gives $7\mid y,v$ and $\pmod{7^3}$ gives $7\mid z,w$. Thus $\left(\frac{x}{7},\frac{y}{7},\frac{z}{7},\frac{u}{7},\frac{v}{7},\frac{w}{7}\right)$ is an interger solution, too. By infinite descent $x=y=z=u=v=w=0$ is the only solution.

Consider this equation $\pmod{7}$. Then, we note that we must have \[x^3 \equiv 2u^3\]which by checking the possible values of $a^3 \pmod{7}$ for integers $a$ gives us that $x\equiv u \equiv 0 \pmod{7}$. Thus, $7\mid x,u$. Setting $x=7x'$ and $u=7u'$ and dividing both sides of the equation by 7 we obtain, \[49x'^3 + y^3 + 7z^3 = 98u'^3 + 2v^3 + 14w^3\]But this is the same equation as before. This implies $7 \mid y,v$ Thus, we set $v=7v'$ and $y=7y'$. Plugging this in and simplifying we obtain \[7x'^3+49y'+z^3 = 14u'^3 + 98v'^3 + 2w^3\]This this also the same as before! Thus, $7 \mid z,w$. We again plug in $z=7z'$ and $w=7w'$ and simplify to obtain, \[x'^3 + 7y'^3 + 49z'^3 = 2u'^3 + 14v'^3 + 98w'^3\]This is exactly the same as the initial equation and we have that $7 \mid \gcd(x,y,z,u,v,w)$. Then, note that this cannot occur infinitely ($v_7\gcd(x,y,z,u,v,w)$) which implies that the only solution is \[(x,y,z,u,v,w)=(0,0,0,0,0,0)\]as desired.

Note that $x^3 \equiv 2u^2 \text{ mod } 7 \implies x \equiv y \equiv 0 \text{ mod } 7$. Now taking $x = 7a, y = 7b, 49a^3+y^3+7z^3=98u^3+2v^3+14w^3.$. Repeating the argument thrice, $x, y, z, u, v, w$ are all multiples of 7. Now descending infinitely, the only solution is $$(x, y, z, u, v, w) = (0, 0, 0, 0, 0, 0)$$.