Maximum of the sum of cube (1) $-2\leq a_{i} \leq 2$ $~(i=1,2,3,4,5)$    
(2) $\displaystyle\sum_{cyclic}a_{i}=0$    
then, find the maximum value of $\displaystyle\sum_{cyclic}a_{i}^{3}$
also, can it be generalized as for $\displaystyle\sum_{cyclic}a_{i}^{n}$ $~(n\in N)$
 A: The function $f(u) = u^3$ is convex for $u \ge 0$ and concave for $u \le 0$.  From this we have for $u, v \in \{ a_i \}$:
$$ f(u) + f(v) \le \begin{cases}
2f\left(\dfrac{u+v}2\right) & \qquad u, v \in [-2,0] \\
f(2) + f(u+v-2) & \qquad  u, v \in [0, 2]
\end{cases} \tag{1}$$
Also, we have, if $ u < 0 < v$: 
$$ f(u) + f(v) \le \begin{cases}
f(0) + f(u+v) & \qquad  u+v  \le 0 \\
f(2) + f(u+v-2) & \qquad  u+v \ge 0
\end{cases} \tag{2}$$
WLOG, we can assume $a_i$ are ordered in descending order.  Let the first $k$ terms be positive.  Then using the reasoning in $(1)$ above we have
$$\sum_{i=1}^k f(a_i) \le (k-1)f(2) + f\left(\sum_{i=1}^k a_i - 2(k-1) \right)$$
Thus it is clear that we can set $k-1$ of these variables to $2$ for maximising the above partial sum.  Let the remaining positive variable be $a_k$.
Similarly, for the terms which are $\le 0$, we have from $(1)$:
$$\sum_{i=k+1}^5 f(a_i) \le (5-k)f\left(\frac1{5-k} \sum_{i=k+1}^5 a_i \right) $$ 
hence we can replace all such terms with their arithmetic mean $\mu$ to maximise this partial sum.
Now we have $\mu < 0 < a_k$ and hence using $(2)$ we can replace $(\mu, a_k)$ with $(2, \mu+a_k-2)$ or $(0, \mu+a_k)$ to increase the sum (depending on the sign of $\mu + a_k$), thereby making all positive terms equal to $2$. So we have that the maximum is when some $k$ of the $a_i$s are $2$, and the remaining are all equal. Obviously $\mu = -\dfrac{2k}{5-k}$, giving a sum of 
$$g(k) = kf(2) - (5-k)f\left(\frac{2k}{5-k}\right) = 8k - \frac{(2k)^3}{(5-k)^2} $$
It is easy to check that $g'(k) = 0$ when $k = \frac53$, and that $k = 2$ gives the maximum among integers in $[1, 5]$. 
The method can be adapted in general for real $a_i \in [c, d]$, and natural $i \le m$ to find the maximum of $\sum a_i^n$ subject to $n \in \mathbb{N}$ and $\sum a_i = 0$.
