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 $x^2+x+1=0$.


Alternate method:
$(x^4 - 1)(x^2 -1)$ has a degree of six, we can write:
$$(x^4 -1)(x^2-1)=Q(x) \left[ 1+x +x^2 \right] + R(x) $$
Evaluate (1) at $\{  \omega, \omega^2 \}$:
$$ \begin{align} R( \omega) &= ( \omega-1)(\omega^2 - 1)= ( \omega -1)( \frac{1}{\omega}-1) \\
 R( \omega^2) &= (\omega^2 -1)(\omega-1) =  ( \omega -1)( \frac{1}{\omega}-1) \end{align} $$
This means that, $$xR(x) -(x-1)(1-x) = (x - \omega)(x- \omega^2)$$
Rearrange for the answer which is $R(x)=3$.
Note: $(x^4 -1)(x^2-1)$ when divided by $1+x+x^2$ must give a polynomial less than degree two. This follows via the remainder. See Wiki.
