B. If real numbers $x, y$ satisfy
$$
x^{3}+y^{3}=26 \text {, and } x y(x+y)=-6 \text {, }
$$

then $|x-y|=$ $\qquad$
B. 4 .

From the given information, we have
$$
\begin{array}{l}
(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)=8 \\
\Rightarrow x+y=2 \Rightarrow x y=-3 . \\
\text { Therefore, }|x-y|=\sqrt{(x+y)^{2}-4 x y}=4 .
\end{array}
$$