Example 24. Solve the equation

$$
\sqrt[3]{x+1}-\sqrt[3]{x-1}=\sqrt[6]{x^{2}-1}
$$
Solution. Solving the inequality $x^{2}-1 \geqslant 0$, we find the domain of the equation (10) - intervals $-\infty < x \leq -1$ and $x \geq 1$; in this case:

if $x < -1$ or $x > 1$, then $\sqrt[3]{x+1}=\sqrt[6]{(x+1)^{2}}$ and $\sqrt[3]{x-1}=\sqrt[6]{(x-1)^{2}}$

Therefore, on this part of the domain, dividing both sides of equation (10) by $\sqrt[6]{x^{2}-1}$, we have an equivalent system of two systems:

$$
\left\{\begin{array} { l } 
{ x < -1 \text{ or } x > 1 \\
\sqrt[6]{\frac{x+1}{x-1}}-\sqrt[6]{\frac{x-1}{x+1}}=1
\end{array}\right.\right.
$$

For each $x$ from the intervals $-\infty < x \leq -1$ and $x \geq 1$, let $\sqrt[6]{\frac{x+1}{x-1}}=t$ and $\sqrt[6]{\frac{x-1}{x+1}}=\frac{1}{t}$.

Considering that

$$
\left\{\begin{array} { l } 
{ t > 0 , } \\
{ - t + 1 / t = 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
t>0, \\
t^{2}+t-1=0
\end{array} \Leftrightarrow t=\frac{\sqrt{5}-1}{2}\right.\right.
$$

and

$$
\left\{\begin{array} { l } 
{ t > 0 } \\
{ t - 1 / t = 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
t>0 \\
t^{2}-t-1=0
\end{array} \Leftrightarrow t=\frac{\sqrt{5}+1}{2}\right.\right.
$$

we have: the system (11) is equivalent to the system

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-067.jpg?height=217&width=1199&top_left_y=2233&top_left_x=428)

i.e., the system

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-067.jpg?height=217&width=1108&top_left_y=2490&top_left_x=474)

## § 2. IRRATIONAL EQUATIONS

We find that $\left(\frac{\sqrt{5}-1}{2}\right)^{6}=9-4 \sqrt{5}$ and $\left(\frac{\sqrt{5}+1}{2}\right)^{6}=$ $=9+4 \sqrt{5}$.

For $c \neq 1$ we have:

$$
\begin{aligned}
\frac{x+1}{x-1}=c \Leftrightarrow x+1=c x-c \Leftrightarrow 1+c= & x(c-1) \Leftrightarrow \\
& \Leftrightarrow x=\frac{c+1}{c-1} \Leftrightarrow x=1+\frac{2}{c-1}
\end{aligned}
$$

From this, we get that the numbers

$$
\begin{aligned}
x_{1}=1+\frac{2}{(9-4 \sqrt{5})-1} & =1+\frac{2}{8-4 \sqrt{5}}=1+\frac{1}{4-2 \sqrt{5}}= \\
=1 & +\frac{4+2 \sqrt{5}}{16-20}=1+\frac{4+2 \sqrt{5}}{-4}=\frac{-\sqrt{5}}{2} \\
x_{2}=1+\frac{2}{(9+4 \sqrt{5})-1} & =1+\frac{2}{8+4 \sqrt{5}}=1+\frac{1}{2 \sqrt{5}+4}= \\
& =1+\frac{2 \sqrt{5}-4}{20-16}=1+\frac{2 \sqrt{5}-4}{4}=\frac{\sqrt{5}}{2}
\end{aligned}
$$

are the roots of the corresponding equations of the last system. Thus, equation (10) is equivalent to the system

$$
\left\{\begin{array} { l } 
{ x < -1 \text{ or } x > 1 \\
x=\sqrt{5} / 2
\end{array}\right.\right.
$$

Therefore, the set of all solutions to equation (10) consists of two numbers: $-\sqrt{5} / 2$ and $\sqrt{5} / 2$.

An equation of the form

$$
\sqrt[2 n]{f(x)}=g(x), \quad n \in \mathbf{N}
$$

is equivalent to the system

$$
\left\{\begin{array}{l}
g(x) \geqslant 0 \\
f(x)=g^{2 n}(x)
\end{array}\right.
$$