Let P(x)=x3+x+1 and Q(x)=x3−x+1.
We are given that when f(x) is divided by P(x), the remainder is 3, so we can write f(x)=P(x)q1(x)+3 for some polynomial q1(x).
When f(x) is divided by Q(x), the remainder is x+1, so we can write f(x)=Q(x)q2(x)+x+1 for some polynomial q2(x).
We want to find the remainder when f(x) is divided by x6+2x3−x2+1.
Note that
\begin{align*} P(x) Q(x) &= (x^3 + x + 1)(x^3 - x + 1) \ &= (x^3 + 1)^2 - x^2 \ &= x^6 + 2x^3 + 1 - x^2 \ &= x^6 + 2x^3 - x^2 + 1. \end{align*}
Let R(x)=x6+2x3−x2+1. Then R(x)=P(x)Q(x).
From f(x)=P(x)q1(x)+3,
\begin{align*} f(x) &= P(x) q_1(x) + 3 \ &= P(x) [Q(x) q_3(x) + r_1(x)] + 3 \ &= R(x) q_3(x) + P(x) r_1(x) + 3 \end{align*}
for some polynomials q3(x) and r1(x), where degr1(x)<3.
From f(x)=Q(x)q2(x)+x+1,
\begin{align*} f(x) &= Q(x) q_2(x) + x+1 \ &= Q(x) [P(x) q_4(x) + r_2(x)] + x+1 \ &= R(x) q_4(x) + Q(x) r_2(x) + x+1 \end{align*}
for some polynomials q4(x) and r2(x), where degr2(x)<3.
Since f(x)≡3(modP(x)) and f(x)≡x+1(modQ(x)), we must have r(x)≡3(modP(x)) and r(x)≡x+1(modQ(x)).
If r(x)=ax2+bx+c, then
\begin{align*} a(x^2+x+1) + b(x+1) + c &\equiv 3 \pmod{x^3+x+1} \ a(x^2-x+1) + b(x+1) + c &\equiv x+1 \pmod{x^3-x+1} \end{align*}
Since r(x) must have degree less than 6, let r(x)=ax5+bx4+cx3+dx2+ex+f.
Since r(x)≡3(modx3+x+1) and r(x)≡x+1(modx3−x+1), we can let r(x)=3.
Then r(2)=3.
Final Answer: The final answer is 3
