Dividing $x^{81}+x^{48}+2x^{27}+x^{6}+3$ by $x^{3}+1$ Letting $f(x)=x^{81}+x^{48}+2x^{27}+x^{6}+3$, and we seek to divide $f(x)$ by $x^{3}+1$, or just any $x^3-a$. Is subbing $y=x^3$ allowed in this kind of division? I've experimented doing so and so far it seems correct.

Even without the substitution, the problem will succumb to
polynomial long division.
Therefore, the only issue is how much perspiration you can replace with inspiration.
As already discussed, setting $y = x^3$ reduces the problem to
$$\text{Divide} ~g(y) = (y^{27} + y^{16}  + 2y^9 + y^2 + 3) ~\text{by}~ (y+1). \tag1$$
In attacking equation (1), note that

*

*$(16 - 9) = (9 - 2)$.


*the middle terms have coefficients : $1,2,1$ which match the 2nd row of Pascal's triangle.
This suggests that the three middle terms of equation (1) above can be re-expressed as
$$(y^2)(y^7 + 1)^2.$$
Further, it is well settled that for $n$ odd, $(y + 1)$ divides $(y^n + 1).$
For example: $(y + 1) \times (y^2 -y + 1) = (y^3 + 1)$ and 
$(y + 1) \times (y^4 -y^3 + y^2 - y + 1) = (y^5 + 1).$
Therefore, you now know (immediately) that $(y + 1)$ divides both $(y^{7} + 1)$ and $(y^{27} + 1).$
Therefore, you can immediately compute the division of $g(y)$ by $(y + 1)$, avoiding polynomial long
division, and immediately conclude that the remainder will be $2$.
