Find the remainder of f divided by g ( x ) = x 4 +...

Let the polynomial be $P(x)$. It is given that for some polynomials $Q(x),Q_1(x)$
$P(x)=(x^2+x+1)Q(x)+1-x$
$P(x)=(x^2-x+1)Q_1(x)+3x+5$
Now it is well known that $x^2+x+1$ has the zeros as $\omega ,{\omega }^2$ So We get
$P(\omega )=1-\omega$
$P({\omega }^2)=1-{\omega }^2$
$P(-\omega )=3\omega +5$
$P(-{\omega }^2)=-3{\omega }^2+5$
Now let us assume for some polynomial $Q_2(x)$ we have
$P(x)=(x^4+x^2+1)Q_2(x)+A{x}^3+B{x}^2+Cx+D$
Using the fact that $x^4+x^2+1$ has the zeros $\omega ,{\omega }^2,-\omega ,-{\omega }^2$ we get four linear equations as:
$\left[\begin{array}{cccc}1& {\omega }^2& \omega & 1\\ 1& \omega & {\omega }^2& 1\\ -1& {\omega }^2& -\omega & 1\\ -1& \omega & -{\omega }^2& 1\end{array}\right]\left[\begin{array}{c}A\\ B\\ C\\ D\end{array}\right]=\left[\begin{array}{c}1-\omega \\ 1-{\omega }^2\\ 5-3\omega \\ 5-3{\omega }^2\end{array}\right]$
Now let us solve by Cramer's rule. The determinant of the matrix is $\mathrm{\Delta }=12$, ${\mathrm{\Delta }}_1=-24$, ${\mathrm{\Delta }}_2=24$, ${\mathrm{\Delta }}_3=12$ and ${\mathrm{\Delta }}_4=60$. Thus the values of A,B,C,D are
$\begin{array}{rl}& A=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=-2\\ & B=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=2\\ & C=\frac{{\mathrm{\Delta }}_3}{\mathrm{\Delta }}=1\\ & D=\frac{{\mathrm{\Delta }}_4}{\mathrm{\Delta }}=5\end{array}$
So the required remainder is
$-2x^3+2x^2+x+5$