3. Given a positive real number $x$ satisfies
$$
x^{3}+x^{-3}+x^{6}+x^{-6}=2754 \text {. }
$$

then $x+\frac{1}{x}=$ $\qquad$
3. 4 .

Transform the given equation into
$$
\begin{array}{l}
\left(x^{3}+x^{-3}\right)^{2}+\left(x^{3}+x^{-3}\right)-2756=0 \\
\Rightarrow\left(x^{3}+x^{-3}+53\right)\left(x^{3}+x^{-3}-52\right)=0 .
\end{array}
$$

Notice that, $x^{3}+x^{-3}+53>0$.
Therefore, $x^{3}+x^{-3}=52$.
Let $b=x+x^{-1}$.
Then $x^{3}+x^{-3}=\left(x+x^{-1}\right)\left[\left(x+x^{-1}\right)^{2}-3\right]$
$$
\begin{array}{l}
\Rightarrow b\left(b^{2}-3\right)=52 \\
\Rightarrow(b-4)\left(b^{2}+4 b+13\right)=0 \\
\Rightarrow b=4 .
\end{array}
$$

Thus, $x+\frac{1}{x}=4$.