Let the numbers be $a_i$, $1\le i\le 9$ with $S=\sum_{1\le i\le 9}a_i$. Set $S-a_i = (2\cdot i)^3$ for $1\le i\le 9$. This ensures (a) the sum of any eight is a cube; and (b) the sum of all, which is $\frac18 \sum_{1\le i\le 9}(2\cdot i)^3 = 45^2$ is a perfect square. Having chosen this, prescribing $a_i$ is easy: simply let $a_i = 45^2 - (2\cdot i)^3$, $1\le i\le 9$ (clearly they are all distinct and non-zero as $45$ is odd).