Let $f(x)$ be a polynomial such that $f(x^2+1)=x^4+x^2+1$. Find $f(x^2-1)$.

Solution

1. Let $u=x^2+1$, so $x^2=u-1$.
2. Then $$\begin{array}{rl}f(u)& =(u-1{)}^2+(u-1)+1\\ & =u^2-2u+1+u-1+1\\ & =u^2-u+1\end{array}$$.
3. We want $f(x^2-1)$, so substitute $u=x^2-1$ into $f(u)=u^2-u+1$.
4. $$\begin{array}{rl}f(x^2-1)& =(x^2-1{)}^2-(x^2-1)+1\\ & =x^4-2x^2+1-x^2+1+1\\ & =x^4-3x^2+3\end{array}$$.
5. Hence the answer is (C).