1. Given real numbers $a, b, c$ satisfy
$$
\sqrt[3]{a-b}+\sqrt[3]{b-c}=2, a-c=26 \text {. }
$$

Then the value of $a-2 b+c$ is ( ).
(A) 26
(B) 28
(C) -28
(D) $\pm 28$
-1. D.
Let $m=\sqrt[3]{a-b}, n=\sqrt[3]{b-c}$. Thus, $m+n=2, m^{3}+n^{3}=a-c=26$.
Solving this, we get $(m, n)=(3,-1)$ or $(-1,3)$.
Therefore, $a-2 b+c=m^{3}-n^{3}= \pm 28$.