115. Solve the equation:
a) $x^{3}-3 x-2=0$;

b) $x^{3}-19 x-30=0$

c) $x^{3}+4 x^{2}+6 x+4=0$.
Solution. a) Factorize the left side of the equation:

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

We have $(x-2)\left(x^{2}+2 x+1\right)=0$, hence $x_{1}=2, x_{2}=x_{3}=-1$.

b) $x^{3}-19 x-30=x^{3}-25 x+6 x-30=x\left(x^{2}-25\right)+6(x-5)=$ $=(x-5)\left(x^{2}+5 x+6\right)$.

We have $(x-5)\left(x^{2}+5 x+6\right)=0$, hence $x_{1}=5, x_{2}=2, x_{3}=3$.

c) $x^{3}+4 x^{2}+6 x+4=x^{3}+2 x^{2}+2 x^{2}+4 x+2 x+4=$ $=x^{2}(x+2)+2 x(x+2)+2(x+2)=(x+2)\left(x^{2}+2 x+2\right)$.

We have $(x+2)\left(x^{2}+2 x+2\right)=0$, hence $x=-2$.

Answer: a) $2,-1$; b) $5,2,3$; c) -2.

$\square$ 116. Solve the equation:
a) $(x-1)^{3}+(2 x+3)^{3}=27 x^{3}+8$;

b) $x^{3}+\left(1-a^{2}\right) x+a=0$;

c) $(a-x)^{3}+(b-x)^{3}=(a+b-2 x)^{3}$.

Solution. a) By expanding the left and right sides of the equation as sums of cubes, we get:

$$
(3 x+2)\left(3 x^{2}+9 x+13\right)=(3 x+2)\left(9 x^{2}-6 x+4\right)
$$

from which

$$
\begin{gathered}
(3 x+2)\left(-6 x^{2}+15 x+9\right)=0,-3(3 x+2)\left(2 x^{2}-5 x-3\right)=0 \\
3 x+2=0, x_{1}=-\frac{2}{3} \\
2 x^{2}-5 x-3=0, x_{2}=-\frac{1}{2}, x_{3}=3
\end{gathered}
$$

b) Transform the left side of the equation:

$$
\begin{aligned}
x^{3}+\left(1-a^{2}\right) x+a= & x^{3}+x-a^{2} x+a=x\left(x^{2}-a^{2}\right)+(x+a)= \\
& =(x+a)\left(x^{2}-a x+1\right)
\end{aligned}
$$

We have $(x+a)\left(x^{2}-a x+1\right)=0$. Solving the equation $x^{2}-a x+1=0$, we find $D=a^{2}-4$. If $D>0$ (i.e., $a>2$ or $a<-2$), then the equation has two roots:

$$
x_{1}=\frac{a-\sqrt{a^{2}-4}}{2} \text { and } x_{2}=\frac{a+\sqrt{a^{2}-4}}{2}
$$

Thus, the given equation has one, two, or three roots depending on $a$.

c) Since $a+b-2 x=(a-x)+(b-x)$, we have the equation

$$
(a-x)^{3}+(b-x)^{3}=((a-x)+(b-x))^{3},
$$

from which

$$
\begin{gathered}
(a-x)^{3}+(b-x)^{3}-(a-x)^{3}-3(a-x)^{2}(b-x)-3(a-x)(b-x)^{2}- \\
-(b-x)^{3}=0
\end{gathered}
$$

That is,

$$
3(a-x)(b-x)(a-x+b-x)=0, \quad 3(a-x)(b-x)(a+b-2 x)=0
$$

By setting each of the three factors containing $x$ to zero, we find the solutions to the given equation.

Answer: a) $-\frac{2}{3},-\frac{1}{2}, 3$; b) the equation has three roots $\frac{a-\sqrt{a^{2}-4}}{2}, \frac{a+\sqrt{a^{2}-4}}{2},-a$, if $a>2$ or $a<-2$; two roots $\frac{a}{2}$ and $-a$, if $a=-2$ or $a=2$; one root, equal to $-a$, if $-2<a<2$; c) $x_{1}=a, x_{2}=b, x_{3}=\frac{a+b}{2}$.