Remainder Theorem Technique 
Determine the remainder when $(x^4-1)(x^2-1)$ is divided by $1 + x + x^2$ (HMMT 2000, Guts Round)

A. Write the division in the form:
$$(x^4-1)(x^2-1)= (1 + x + x^2)Q(x) + R(x)$$
B. Multiply both sides by $x-1$:
$$(x-1)(x^4-1)(x^2-1)= (x^3-1)Q(x) + R(x)(x-1)$$
C. Substitute $x^3=1,x\neq1$, and reduce the resulting equation:
$$(x-1)(x-1)(x^2-1)= R(x)(x-1)$$
D. Divide both sides by $x-1$:
$$R(x)=(x-1)(x^2-1)=x^3 -x - x^2 + 1=-(x^2+x+1)+3=3$$

For someone who knows the method, is it valid to skip Steps B and D, directly substitute $x^3=1,x\neq1$ and use the fact that $x$ is a cube root of unity to get $x^2+x+1=0$.


Yes, you can skip those steps.
The verbose equivalent is that $$\begin{align}(x^4-1)(x^2-1)&=(x^4-x)(x^2-1)+(x-1)(x^2-1)\\
&=(x^2+x+1)x(x-1)(x^2-1)+(x-1)(x^2-1)
\end{align} $$
So $(x-1)(x^2-1)$ has the same remainder as $(x^4-1)(x^2-1)$ when dividing by $x^2+x+1.$

Answering the question in comments, yes, you can replace $x^2+x+1$ with zero.
Here, you can write:
$$(x-1)(x^2-1)=(x-1)(x^2+x+1)-(x-1)(x+2)$$
So you get the same remainder for $-(x-1)(x+2)$ as $(x-1)(x^2-1).$
