Is it possible to factor expression $x^m+x^n+c$ into a product of two or more terms? I apologize ahead of time that my question will be hazy, but I'm very much under an impression that it has a solution, or possibly even a class of solutions.  Here, goes...
Long, long time ago, in middle school I recall we were given an equation like so:
$$x^m+x^n+c$$
where $m$, $n$, and $c$ were specific integer numbers.  I am pretty sure each one of them was below 10, and $c$ was most definitely $1$
The task was to factor it out into a parenthesized expression, aka a product of two or more parenthesized terms.
I had no clue how to do that.  But later, someone showed me a trick that made it possible to try.  The trick was to add and then subtract $1$, and after that, it was more clear how to group and parenthesize it.  I remember questioning this trick or why it was needed but somehow it helped.
For example, if $m=2$, $n=3$ and $c=-3$:
$$x^2+x^3-3$$
$$=x^2+x^3-3+1-1$$
$$=x^2+x^3+1-4$$
$$=(x^3+1)+(x^2-4)$$
$$=(x+1)(x^2-x+1)+(x-2)(x+2)$$
$$=(x+1)(x^2-x+1)+(x-2)(x+1+1)$$
$$=(x+1)(x^2-x+1)+(x-2)(x+1)+(x-2)$$
$$=(x+1)(x^2-1)+(x-2)$$
etc.... I am not sure in this case it's possible to form a product of parenthesized expressions.  
My question is
Does such a such factorization exist for some numbers $m, n, c$?
And, as a side-note, what is this trick with adding and subtracting one and why did that help?
In my memory the equation looked something like $x^7+x^3+1$, but I am fairly sure that was not the case, the numbers were likely different.

For naturals $m$ and $n$ we obtain that 
$x^{3m-1}+x^{3n-2}+1$ is divisible by $x^2+x+1$ 
because $$x^{3m-1}+x^{3n-2}+1=x^{3m-1}-x^2+x^{3n-2}-x+x^2+x+1=$$
$$=\left((x^3)^{m-1}-1\right)x^2+\left((x^3)^{n-1}-1\right)x+x^2+x+1.$$
Here $$\mathbb N=\{1,2,3,...\}$$
