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.
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1Yes, it is allowed and indeed is sensible if every power of $x$ is a multiple of $3$. You can undo the substitution at the end (if you could not undo it, then that might suggest it was not a good idea) – Henry Dec 21 '20 at 02:52
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Of course it is allowed. Since the substitution doesn't change the roots of the polynomials, and thus the divisibility.
In fact, any substitution $y=g(x)$, where $g(x)$ is a polynomial can be applied.
FFjet
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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 $2$.
user2661923
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