Find all integral solutions of the equation $x^n+y^n+z^n=2016$ 
Find all integral solutions of equation
  $$x^n+y^n+z^n=2016,$$
  where $x,y,z,n -$ integers and $n\ge 2$

My work so far:
1) $n=2$ $$x^2+y^2+z^2=2016$$
I used wolframalpha n=2 and I received the answer to the problem (Number of integer solutions: 144)
2) $n=3$ 
I used wolframalpha n=3 and I not received the answer to the problem 
How to do without wolframalpha?
 A: The case $n=3$ has been studied intensively. Since $2016\equiv 0\bmod 9$ there should be solutions (probably even infinitely many) of $x^3+y^3+z^3=2016$, but many of them  will be very large.
For a reference here see this MO-question. For example, the smallest solution to $x^3+y^3+z^3=30$ is $(x,y,z)=(−283059965,−2218888517,2220422932)$, and for $x^3+y^3+z^3=33$ the status is already unknown.
A: Here an incomplete answer about the possibilities for $2016$. It is clear that the problem would become more difficult for higher numbers.
$$\boxed {n=2}$$
It is known enough that an integer $n$ is a sum of three squares if and only if $n$ is not of the form $4^a(8b+7)$ so because $2016=4^2(8\cdot15+6)$ we can ensure that 
$x^2+y^2+z^2=2016$ has solutions. Calculation gives for integer non negatives
 $$(x,y,z)\in\{(4,8,44),(4,20,40),(12,24,36)\}$$
The other solutions are obtained trivially by permutations and change of signs so we get 
the $3\cdot8\cdot6=144$ solutions the OP said given by Wolphram.
$$\boxed {n=3}$$ The actual state of the question has been  well explained by Dietrich Burde above.
$$\boxed {n=4}$$
We try to see this case in some detail. Note that $6^4=1296<2016<7^4=2401$ so we have to see the $4$-th powers   $0,1,16,81,256,625,1296$ and see at the equations
$$\begin{cases}y=\sqrt[4]{2016-z^4}\\y=\sqrt[4]{2015-z^4}\\ y=\sqrt[4]{2000-z^4}\\ y=\sqrt[4]{1935-z^4}\\ y=\sqrt[4]{1760-z^4}\\ y=\sqrt[4]{1391-z^4}\\ y=\sqrt[4]{720-z^4 }\end{cases}$$ where the numbers are $2016-x^4$ corresponding to $x=0,1,2,3,4,5,6$ respectively. There are no solutions (excepting some mistake in calculations).
$$\boxed {n=6,8,…2k}$$
Since $3^6\lt 2016\lt 4^6$ and $2^8\lt 2016\lt 3^8$ and $2^{10}\lt 2016\lt 2^{11}$ it is easy to examine the possibilities (apparently no solutions).
$$\boxed {n=2k+1}$$
It seems that this problem is similar to that in which $n = 3$ due to the appearance of negative integers.
A: An approach that can sometimes help to find solutions to this type of equation is to consider the prime factors of the given integer.  In this case:
$$2016 = 2^5.3^2.7 = 2^5(2^6-1) = 2^5((2.2^5)-1)$$
Hence:
$$\mathbf{2016 = 4^5 + 4^5 + (-2)^5}$$
And also (finding a solution of $x^3+y^3+z^3=252$ by trial and error or from this list):
$$2016 = 2^3.252 = 2^3(7^3 - 3^3 - 4^3)$$
Hence:
$$\mathbf{2016 = 14^3 + (-6)^3 + (-8)^3}$$
