137. Knowing that $\cos ^{6} x+\sin ^{6} x=a$, find $\cos ^{4} x+\sin ^{4} x$.
S o l u t i o n. Transform the expression $\cos ^{4} x+\sin ^{4} x$ :

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

Find the value of the product $\cos ^{2} x \sin ^{2} x$ from the given equation, for which we transform its left side:

$$
\begin{gathered}
\cos ^{6} x+\sin ^{6} x=\left(\cos ^{2} x\right)^{3}+\left(\sin ^{2} x\right)^{3}= \\
=\left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{4} x-\cos ^{2} x \sin ^{2} x+\sin ^{4} x\right)= \\
=\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-3 \cos ^{2} x \sin ^{2} x=1-3 \cos ^{2} x \sin ^{2} x
\end{gathered}
$$

We have $1-3 \cos ^{2} x \sin ^{2} x=a$, from which $\cos ^{2} x \sin ^{2} x=\frac{1-a}{3}$. Therefore,

$$
\cos ^{4} x+\sin ^{4} x=1-2 \cos ^{2} x \sin ^{2} x=1-2 \cdot \frac{1-a}{3}=\frac{1+2 a}{3}
$$

A n s w e r: $\frac{1+2 a}{3}$.