5. In $\triangle A B C$, it is known that
$$
A B=A C=1, B C=x, \angle A=36^{\circ} \text {. }
$$

Then $\sqrt[3]{x-x^{2}}=(\quad)$.
(A) $\frac{\sqrt{5}}{2}$.
(B) $\frac{3-\sqrt{5}}{2}$
(C) 1
(D) $\frac{\sqrt{5}-1}{2}$
5. D.

As shown in Figure 8, construct the angle bisector of $\angle A B C$ intersecting $A C$ at point $D$. Then
$$
\begin{array}{l} 
A D=B D=B C=x, \\
C D=1-x, \\
\text { and } \triangle B C D \backsim \triangle A B C \\
\Rightarrow \frac{B C}{A B}=\frac{C D}{B C} \\
\Rightarrow \frac{x}{1}=\frac{1-x}{x} \\
\Rightarrow x^{2}+x-1=0 \Rightarrow x=\frac{\sqrt{5}-1}{2} . \\
\text { Therefore, } \sqrt[3]{x-x^{2}}=\sqrt[3]{x^{3}}=x=\frac{\sqrt{5}-1}{2} .
\end{array}
$$