3.293. $\cos ^{2} \alpha+\cos ^{2} \beta-2 \cos \alpha \cos \beta \cos (\alpha-\beta)$.
## Solution.

$$
\begin{aligned}
& \cos ^{2} \alpha+\cos ^{2} \beta-2 \cos \alpha \cos \beta \cos (\alpha-\beta)= \\
& =\frac{1+\cos 2 \alpha}{2}+\frac{1+\cos 2 \beta}{2}-2 \cos \alpha \cos \beta \cos (\alpha-\beta)=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{1}{2}(2+(\cos 2 \alpha+\cos 2 \beta))-2 \cos \alpha \cos \beta \cos (\alpha-\beta)= \\
& =\left[\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}\right]= \\
& =\frac{1}{2}(2+2 \cos (\alpha+\beta) \cos (\alpha-\beta))-2 \cos \alpha \cos \beta \cos (\alpha-\beta)= \\
& =1+\cos (\alpha+\beta) \cos (\alpha-\beta)-2 \cos \alpha \cos \beta \cos (\alpha-\beta)= \\
& =1+\cos (\alpha-\beta)(\cos (\alpha+\beta)-2 \cos \alpha \cos \beta)= \\
& =\left[\cos x \cos y=\frac{1}{2}(\cos (x-y)+\cos (x+y))\right]= \\
& =1+\cos (\alpha-\beta)(\cos (\alpha+\beta)-\cos (\alpha-\beta)-\cos (\alpha+\beta))= \\
& =1+\cos (\alpha-\beta)(-\cos (\alpha-\beta))=1-\cos ^{2}(\alpha-\beta)=\sin ^{2}(\alpha-\beta)
\end{aligned}
$$

Answer: $\sin ^{2}(\alpha-\beta)$.