<s>[INST] Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$. 
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.

My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$, then do the same for $x^2-x+1$ and then use that along with the known remainder of $f$ divided by $x^2+x+1$ and $x^2-x+1$ to obtain $A$, $B$, $C$, $D$. However, $Ax^3$ is in the way so I don't know how to proceed nor do I have any other ideas to start with.
 [/INST] Hint:
As the roots of $x^2+x+1=0$ are $w,w^2$ where $w$ is a complex cube root of unity,
the roots of $x^2-x+1=0$ are $-w,-w^2$
we can write $$f(x)$$
$$=p(x)(x^2+x+1)(x^2-x+1)+A(x-w)(x-w^2)(x+w)+B(x-w)(x-w^2)(x+w^2)+C(x-w^2)(x+w)(x+w^2)+D(x-w)(x+w)(x+w^2)$$
$$=p(x)(x^2+x+1)(x^2-x+1)+(x^2-x+1)(c(x-w^2)+d(x-w))+\cdots$$
$$-w+1=f(w)=(-2w)c(w-w^2)\implies2c=-w$$
Similarly, we can find $a,b,d$
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