Solve the following equations and verify the correctness of the obtained roots:
а) $\frac{3}{1-6 x}=\frac{2}{6 x+1}-\frac{8+9 x}{36 x^{2}-1}$,
b) $\frac{3}{1-z^{2}}=\frac{2}{(1+z)^{2}}-\frac{5}{(1-z)^{2}}$.
a) Our equation can also be written as

$$
-\frac{3}{6 x-1}=\frac{2}{6 x+1}-\frac{8+9 x}{(6 x+1)(6 x-1)}
$$

Assuming that $x \neq \frac{1}{6}, x \neq-\frac{1}{6}$

$$
-3(6 z+1)=2(6 x-1)-8-9 x
$$

after rearranging

$$
-21 x=-7
$$

from which

$$
x=\frac{7}{21}=\frac{1}{3}
$$

Verification: The left side

$$
\frac{3}{1-6 \cdot \frac{1}{3}}=\frac{3}{1-2}=\frac{3}{-1}=-3
$$

The right side

$$
\frac{2}{6 \cdot \frac{1}{3}+1}-\frac{8+9 \cdot \frac{1}{3}}{36 \cdot \frac{1}{9}-1}=\frac{2}{2+1}-\frac{8+3}{4-1}=\frac{2}{3}-\frac{11}{3}=-\frac{9}{3}=-3
$$

Thus, the root we obtained indeed satisfies our equation.

b) Assuming that $z \neq 1, z \neq-1$, and considering that $1-z^{2}=(1+z)(1-z)$, our equation can also be written as

$$
3(1+z)(1-z)=2(1-z)^{2}-5(1+z)^{2}
$$

or

$$
3-3 z^{2}=2-4 z+2 z^{2}-5-10 z-5 z^{2}
$$

which simplifies to

$$
14 z=-6
$$

from which

$$
z=-\frac{6}{14}=-\frac{3}{7}
$$

Verification:

Left side:

$$
\frac{3}{1-\frac{9}{49}}=\frac{3 \cdot 49}{49-9}=\frac{147}{40}
$$

Right side:

$$
\begin{aligned}
\frac{2}{\left(\frac{4}{7}\right)^{2}} & -\frac{5}{\left(\frac{10}{7}\right)^{2}}=\frac{2 \cdot 49}{16}-\frac{5 \cdot 49}{100}=\frac{49}{8}-\frac{49}{20}= \\
& =\frac{5 \cdot 49-2 \cdot 49}{40}=\frac{3 \cdot 49}{40}=\frac{147}{40}
\end{aligned}
$$

Thus, the root we obtained indeed satisfies our equation.