3. The number of ordered pairs $(a, b)$ that satisfy $(a+b \mathrm{i})^{6}=a-b \mathrm{i}(a, b \in \mathbf{R})$ is $\qquad$

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3. 8 .

Let $z=a+b$ i. Then
$$
z^{6}=\bar{z} \Rightarrow|z|^{6}=|\bar{z}|=|z| \text {. }
$$

Thus, $|z|=0$ or 1.
When $|z|=0$, $z=0$, i.e., $a=b=0$. Clearly, $(0,0)$ satisfies the condition;
When $|z|=1$, from $z^{6}=\bar{z}$, we know
$$
z^{7}=z \cdot \bar{z}=|z|^{2}=1 \text {. }
$$

Note that, $z^{7}=1$ has 7 complex solutions, meaning there are 7 ordered pairs of real numbers $(a, b)$ that satisfy the condition.

In summary, the number of ordered pairs of real numbers $(a, b)$ that satisfy the condition is 8.