What is the value of $a+b+c$? 
What is the value of $a+b+c$?
if $$a^4+b^4+c^4=32$$
$$a^5+b^5+c^5=186$$
$$a^6+b^6+c^6=803$$

How to approach this kind of problem. Any help.
UPDATE: Thank you all for answers. Now I realize there is no integer solution. But is there any real number solution? I am curious to know.
 A: If $x$ is an integer then $x^4$ and $x^6$ have the same parity. Therefore $a^4+b^4+c^4$ and $a^6+b^6+c^6$ are either both odd or both even.
Therefore there is no solution.
A: Note that $|n|\ge 3 \implies n^4\ge 81.$ Thus in the first equation it is $a,b,c\in\{0,\pm 1,\pm 2\}.$ It is easy to check that the solution is $a=0,b=\pm 2,c=\pm 2,$ or any permutation.
Now, $$a^5+b^5+c^5\le 0^5+2^5+2^5=64<186,$$ from where you can conclude that there is no solution.
A: The lowest possible values for fourth powers of integers are $$0^4 = 0 \qquad 1^4 = (-1)^4 = 1 \qquad 2^4 = (-2)^4 = 16$$
By inspection, it becomes very clear that two of $a,b,c$ must have absolute value $2$, and one must be zero, in order for your first equation to hold. 
Without loss of generality, suppose $c=0$ and $a = \pm b = \pm 2$.
Now $a^5 + b^5$ must be $2^5 + 2^5$, $-2^5 + 2^5$, or $-2^5 -2^5$. None of these is equal to $186$, so there are no solutions in the integers.
A: Let $m = \max\{|a|,|b|,|c|\}$ then the first equation gives us$$m^4 \leq 32\to m\leq 32^{1/4}$$ and the third equation gives us $$803 = a^6+b^6+c^6 \leq 3\cdot m^6 \leq 3(32)^{6/4}$$
which is a contradiction since $3(32)^{6/4} \approx 543.06 < 803$ so there are no real solutions.
A: If you have a system of equations of the form
$$a^{n_1}+b^{n_1}+c^{n_1}=C_1$$
$$a^{n_2}+b^{n_2}+c^{n_2}=C_2$$
$$a^{n_3}+b^{n_3}+c^{n_3}=C_3$$
The first thing you should do is write the first equation as a function of one of the variables. Here, it could be
$$c=\sqrt[n_1]{C-a^{n_1}-b^{n_1}}$$
Substitute that into the second equation:
$$a^{n_2}+b^{n_2}+(\sqrt[n_1]{C-a^{n_1}-b^{n_1}})^{n_2}=C_2$$
Now do some algebra and try to write this equation as a function of $b$, insert that and the equation for $c$ into the last equation, and solve for $a$. You can then solve for $b$; knowing both, you can solve for $c$.
A: Let $E_n=a^n+b^n$, then we have that $E_n=(a+b)E_{n-1}-abE_{n-2}$.
Given your problem, you can put c any value you want, and let n=6, then you will get an easy eqation were you can put one of the two, a or b, and from the equation you will get the other one for a certain c.
