$$a^3 + \cdots + (a+7)^3 = 8a^3 + 84a^2 + 420a + 784$$
So the cubic curve of interest is $8a^3 + 84a^2 + 420a + 784 = b^3$ over $\Bbb Q$. This obviously have a rational point, take for example $(a, b) = (-4, -4)$. Thus it's isomorphic to an elliptic curve.
In particular, use the parametrization
$$(x, y) = \left (\frac{1008b}{2a+7}, \frac{254016}{2a + 7} \right )$$
To reduce the above into the well-known Mordell form
$$\mathbf {E} : y^2 = x^3 -1024192512$$
I do hope this has rank $1$, though. (I believe BSD)
First, the trivial solutions over $\Bbb Q$ include $(-7/2, 0), (-4, -4), (-3, 4)$, the first being simply a real zero of the cubic. Now about computing the torsion, one can try a relatively easy hands-on derivation by either Nagell-Lutz or Mazur (I am not sure which one is easier here). Precisely, it's $\Bbb Z/2\Bbb Z$ so the group of rational points over the elliptic curve is really $\Bbb Z \times \Bbb Z/ 2\Bbb Z$.
The extra solutions may come from the torsion points. It seems that a generator for $\mathbf E$ is $(2016, 84672)$ and from this comes the corresponding integral point $(-2, 6)$.
I am missing something here.
EDIT : Darn, I forgot the negative $(2016, -84672)$. Of course, plugging this in gives $a = -5$ and $b = 6$, and thus all the integer points (and as well rational points, which are the multiplies of the generator precisely).