3. Let $s$ be a non-real root of the equation $x^{7}=1$. Then
\[
\begin{array}{l}
s(s+1)\left(s^{2}-s+1\right)\left(s^{2}+s+1\right) \\
=
\end{array}
\]
3. -1 .

Notice,
$$
\begin{array}{l}
x^{7}-1=(x-1)\left(x^{6}+x^{5}+\cdots+x+1\right) . \\
\text { And } x^{6}+x^{5}+\cdots+x+1 \\
=x^{4}\left(x+\frac{1}{2}\right)^{2}+\frac{1}{2} x^{2}(x+1)^{2}+ \\
\quad\left(\frac{1}{2} x+1\right)^{2}+\frac{1}{4} x^{2}\left(x^{2}+1\right) \\
>0,
\end{array}
$$

Therefore, the equation has only one real root $x=1$.
$$
\begin{array}{l}
\text { Also } x^{6}+x^{5}+\cdots+x+1 \\
=x(x+1)\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)+1,
\end{array}
$$
$$
\begin{array}{l}
\text { Then } s(s+1)\left(s^{2}-s+1\right)\left(s^{2}+s+1\right)+1=0 \\
\Leftrightarrow s(s+1)\left(s^{2}-s+1\right)\left(s^{2}+s+1\right)=-1
\end{array}
$$