3. Given that the real number $x$ satisfies $x^{3}+\frac{1}{x^{3}}=18$. Then $\{x\}+\left\{\frac{1}{x}\right\}=(\quad)$.
(A) $\frac{1}{2}$
(B) $3-\sqrt{5}$
(C) $\frac{3-\sqrt{5}}{2}$
(D) 1
$(2014$, National Junior High School Mathematics League)
Notice,
$$
x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left[\left(x+\frac{1}{x}\right)^{2}-3\right]=18
$$

Let $x+\frac{1}{x}=a$. Then
$$
a\left(a^{2}-3\right)=18 \Rightarrow a=3 \text {. }
$$

Solving gives $x=\frac{3 \pm \sqrt{5}}{2}$.
Clearly, $\{x\}+\left\{\frac{1}{x}\right\}=1$. Therefore, the answer is (D).