## Task B-3.1.

If $\sin x-\cos x=\frac{\sqrt{3}}{3}$, calculate $\sin ^{6} x+\cos ^{6} x$.
## Solution.

If we square the given equality, we get

$$
\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x=\frac{1}{3}
$$

which means $\sin x \cos x=\frac{1}{3}$.

The expression $\sin ^{6} x+\cos ^{6} x$ can be written in the form of a product

$$
\begin{aligned}
\sin ^{6} x+\cos ^{6} x & =\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right) & & 1 \text { point } \\
& =\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x & & 1 \text { point } \\
& =\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x=1-3 \sin ^{2} x \cos ^{2} x . & & 1 \text { point }
\end{aligned}
$$

Using $\sin x \cos x=\frac{1}{3}$, we finally get

$$
\sin ^{6} x+\cos ^{6} x=1-3(\sin x \cos x)^{2}=1-3\left(\frac{1}{3}\right)^{2}=\frac{2}{3}
$$