3.170. Find $\sin ^{4} \alpha+\cos ^{4} \alpha$, if it is known that $\sin \alpha-\cos \alpha=\frac{1}{2}$.
## Solution.

$$
\sin ^{4} \alpha+\cos ^{4} \alpha=\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-2 \sin ^{2} \alpha \cos ^{2} \alpha=1-2 \sin ^{2} \alpha \cos ^{2} \alpha
$$

By squaring both sides of the equation $\sin \alpha-\cos \alpha=\frac{1}{2}$, we get

$$
\sin ^{2} \alpha-2 \sin \alpha \cos \alpha+\cos ^{2} \alpha=\frac{1}{4}
$$

from which

$$
\sin \alpha \cos \alpha=\frac{3}{8}, \sin ^{2} \alpha \cos ^{2} \alpha=\frac{9}{64}
$$

Substituting this value of $\sin ^{2} \alpha \cos ^{2} \alpha$ into the original equation, we get

$\sin ^{4} \alpha+\cos ^{4} \alpha=1-2 \cdot \frac{9}{64}=1-\frac{9}{32}=\frac{32-9}{32}=\frac{23}{32}$.

Answer: $\frac{23}{32}$.