3.467 Given: $\cos 2 \alpha=m$. Find $A=\cos ^{8} \alpha-\sin ^{8} \alpha$.
## Solution.

$A=\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right) \cdot\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right) \cdot\left(\cos ^{4} \alpha+\sin ^{4} \alpha\right)=\cos 2 \alpha \cdot\left(\cos ^{4} \alpha+\right.$ $\left.+2 \cos ^{2} \alpha \cdot \sin ^{2} \alpha+\sin ^{4} \alpha-2 \cos ^{2} \alpha \cdot \sin ^{2} \alpha\right)=\cos 2 \alpha \cdot\left(1-2 \cos ^{2} \alpha \cdot \sin ^{2} \alpha\right)=$ $=\cos 2 \alpha \cdot\left(1-\frac{\sin ^{2} 2 \alpha}{2}\right)=\frac{m\left(1+m^{2}\right)}{2}$.

Answer: $A=\frac{m\left(1+m^{2}\right)}{2}$.