$17 \cdot 75$ As shown in the figure, in $\triangle A B C$, $D$ is on $A C$. $F$ is on $B C$, and $A B \perp A C, A F \perp B C, B D=D C=$ $F C=1 . A C$ should be
(A) $\sqrt{2}$.
(B) $\sqrt{3}$.
(C) $\sqrt[3]{2}$.
(D) $\sqrt[3]{3}$.
(E) $\sqrt[4]{3}$.
[Solution 1] Let $AC = x, \angle DCF = \theta$.
In $\triangle DBC$, since $BD = DC = 1$, we have $\angle CBD = \theta$.
By the exterior angle theorem, $\angle ADB = 2\theta$.
In the right triangle $AFC$, $\cos \angle DCF = \cos \theta = \frac{1}{x}$.
In the right triangle $ABD$, $AD = x - 1$, so
$$
\cos \angle ADB = \cos 2\theta = \frac{x - 1}{1} = x - 1.
$$

By the trigonometric identity $\cos 2\theta = 2 \cos^2 \theta - 1$,
we have $2\left(\frac{1}{x}\right)^2 - 1 = x - 1$,
which simplifies to $2 - x^2 = x^3 - x^2$,
solving for $x$ gives $x = \sqrt[3]{2}$.
Therefore, the answer is $(C)$.
[Solution 2] Draw $DG \perp BC$ at $G$, and let $AC = x, GC = y$.

Since $\triangle BDC$ is an isosceles triangle and $DG$ is the altitude to $BC$, we have $BC = 2y$.
By the similarity of triangles $\triangle DCG \sim \triangle ACF \sim \triangle BCA$,

we have $\frac{1}{y} = \frac{x}{1} = \frac{2y}{x}$.
Thus, $y = \frac{1}{x}$ and $y = \frac{x^2}{2}$, eliminating $y$ gives $x^3 = 2$, hence $x = \sqrt[3]{2}$. Therefore, the answer is $(C)$.