<s>[INST] Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $\mathbb{Z}$ Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $\mathbb{Z}$

First I tried to transform this equation, substituting $x = 8-y-z$. So I end up with:
$$x^3 + y^3 + z^3 = 8$$
$$(8-y-z)^3 + y^3 + z^3 = 8$$
Using Wolfram Alpha I expanded this equation and tried to factorize it so finally I got:
$$(z-8)(y^2 + y(z-8) - 8z) = 168$$
$z \in \mathbb{Z}$, which implies $z-8$ is an integer implying that the second term is also an integer.
168 has 16 positive divisors and 16 negative, which means there are 32 distinct cases to be looked, which is a painful work and made me abandon this method in search of a better and easier method.

Second method I tried is Newton Identities.
We want to find the values for the roots $x,y,z$ of the cubic function: $f(x) = x^3 + bx^2 + cx + d$
From the condition we have the following equation:
$$x + y + z = 8 = s_1\text{   and   } x^3 + y^3 + z^3 = 8 = s_3$$
Using Newton Identities we obtain the following relations:
$$s_1 + b = 0 \implies b = -8$$
$$s_2 + bs_1 + 2c = 0$$
$$s_2 - 64 + 2c = 0 \implies c = \frac{64 - s_2}{2}$$
$$s_3 + bs_2 + cs_1 + 3d = 0$$
$$8 - 8s_2 + \frac{64 - s_2}{2} \times 8 + 3d = 0$$
$$8 - 8s_2 + 256 - 4s_2 + 3d = 0$$
$$3d - 12s_2 + 264 = 0$$
$$d - 4s_2 + 88 = 0 \implies s_2 = \frac{88+d}{4}$$
Substituting back we have:
$$c = \frac{64 - s_2}{2} = \frac{64 - \frac{88+d}{4}}{2} = \frac{\frac{256 - 88 - d}{4}}{2} = \frac{168 - d}{8}$$
And I'm stuck here, because I know only the ratio between $c$ and $d$, which is a consequence of not knowing the value $s_2 = x^2 + y^2 + z^2$
 [/INST] Hint: Calculate $(x-8)(y-8)(z-8)$.

There are several ways to do the calculation. Working from the last equation in the first method, observe that the second term can be factorized into $(y-8)(y+z)= -(y-8)(x-8)$, which gives us

 $(x-8)(y-8)(z-8) = -168 $

Use this, along with the factors of 168 and that $x-8+y-8+z-8 = -16$ to greatly restrict the cases you have to look at. For example, as Mark comments, since the sum of 3 numbers is even, so either one of them is even, all all three of them are. 

Now, if you don't want an 'Wolfram Alpha' expansion of the terms, you could do the expansion yourself, it's not that hard. The way that I arrived at the calculation used the fact that $a^3 + b^3 + c^3 - 3abc = (a+b+c)( (a+b+c)^2 - 3ab -3bc - 3ca) $
Note that since you want to solve this as a Diophantine equation, using algebraic methods can only get you so far. At some point in time, you must change it into a 'number theoretic way', like analyzing the factors, or gcd of terms.
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