1. Given the identity
$$
\begin{array}{l}
x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4} \\
=(x+1)^{4}+b_{1}(x+1)^{3}+b_{2}(x+1)^{2}+ \\
\quad b_{3}(x+1)+b_{4} .
\end{array}
$$

Express $b_{3}$ in terms of $a_{1}, a_{2}, a_{3}, a_{4}$, then $b_{3}=$
$\qquad$
$$
-1 .-4+3 a_{1}-2 a_{2}+a_{3} \text {. }
$$

Substituting $x=-1$ into the known identity, we get
$$
b_{4}=1-a_{1}+a_{2}-a_{3}+a_{4} .
$$

Rearranging and simplifying, we have
$$
\begin{array}{l}
\left(x^{4}-1\right)+a_{1}\left(x^{3}+1\right)+a_{2}\left(x^{2}-1\right)+a_{3}(x+1) \\
=(x+1)^{4}+b_{1}(x+1)^{3}+b_{2}(x+1)^{2}+b_{3}(x+1),
\end{array}
$$

which is
$$
\begin{array}{l}
(x+1)\left[(x-1)\left(x^{2}+1\right)+\right. \\
\left.a_{1}\left(x^{2}-x+1\right)+a_{2}(x-1)+a_{3}\right] \\
=(x+1)\left[(x+1)^{3}+b_{1}(x+1)^{2}+\right. \\
\left.b_{2}(x+1)+b_{3}\right] . \\
\text { Hence }(x-1)\left(x^{2}+1\right)+a_{1}\left(x^{2}-x+1\right)+a_{2}(x-1)+a_{3} \\
=(x+1)^{3}+b_{1}(x+1)^{2}+b_{2}(x+1)+b_{3} .
\end{array}
$$

Substituting $x=-1$ again, we get
$$
b_{3}=-4+3 a_{1}-2 a_{2}+a_{3} .
$$