2.15. Solve the equation $\sqrt[3]{1-x}+\sqrt[3]{1+x}=p$, where $p$ is an arbitrary real number.

## 2.4. Different Equations

### 2.16. Solve the equation

$$
|x+1|-|x|+3|x-1|-2|x-2|=x+2
$$
2.15. Let's raise both sides of the equation to the third power:

$$
2+3 \sqrt[3]{1-x^{2}}(\sqrt[3]{1-x}+\sqrt[3]{1+x})=p^{3}
$$

Then substitute $p$ for $\sqrt[3]{1-x}+\sqrt[3]{1+x}$. As a result, we get the equation $2+3 p \sqrt[3]{1-x^{2}}=p^{3}$, from which

$$
x= \pm \sqrt{1-\left(\frac{p^{3}-2}{3 p}\right)^{3}}
$$

This formula makes sense for $p=-1$ and $0<p \leqslant 2$. However, in the process, we applied non-equivalent transformations: we could not lose roots, but we could gain extraneous roots.

Let's analyze our transformations in more detail. Let $u=\sqrt[3]{1-x}, v=\sqrt[3]{1+x}$. Raise both sides of the equation $u+v=p$ to the third power (this is an equivalent transformation). As a result, we get $u^{3}+v^{3}+3 u v(u+v)=p^{3}$. Substitute $p$ for $u+v$. As a result, we get $u^{3}+v^{3}+3 u v p=p^{3}$, i.e., $(u+v)^{3}-p^{3}-3 u v(u+v-p)=0$. Write $(u+v)^{3}-p^{3}$ using the formula for the difference of cubes: $a^{3}-b^{3}=$ $=(a-b)\left(a^{2}+a b+b^{2}\right)$. Factoring out $(u+v-p)$, we get the equation

$$
(u+v-p)\left(u^{2}+v^{2}+p^{2}+u p+v p-u v\right)=0
$$

Extraneous roots can only be associated with the equation $\left(u^{2}+v^{2}+p^{2}+u p+v p-u v\right)=0$, i.e., $(u+p)^{2}+(v+p)^{2}+(u-v)^{2}=0$. Therefore, extraneous roots can only occur if $u=v=-p$, i.e., $\sqrt[3]{1-x}=\sqrt[3]{1+x}=-p$. This system of equations has a unique solution $x=0, p=-1$. The root $x=0$ for $p=-1$ is indeed extraneous.