# Task 2.

Maximum 15 points

Solve the equation:

$$
|3 x+3|^{3}+|3 x-5|^{3}=131
$$

#
# Solution:

Let's make a variable substitution:

$$
t=\frac{3 x+3+3 x-5}{2}=3 x-1
$$

The equation becomes:

$$
|t+4|^{3}+|t-4|^{3}=131
$$

It is not hard to see that the function $f(t)=|t+4|^{3}+|t-4|^{3}$ is even, so it is sufficient to solve the equation for $t \geq 0$. Consider the cases for removing the absolute value:

a) For $0 \leq t \leq 4$, we have

$$
(t+4)^{3}-(t-4)^{3}=131
$$

From this, we find $t=\frac{\sqrt{2}}{4}$.

b) For $t>4$, we have

$$
(t+4)^{3}+(t-4)^{3}=131 \Leftrightarrow 2 t^{3}+96 t-131=0
$$

The function $g(t)=2 t^{3}+96 t-131$ is monotonically increasing, so for $t>4 f(t)>381$, which means there are no solutions in this interval.

We have the solutions $t= \pm \frac{\sqrt{2}}{4}$. From this, we get

$$
x=\frac{t+1}{3}=\frac{ \pm \frac{\sqrt{2}}{4}+1}{3}=\frac{ \pm \sqrt{2}+4}{12}
$$

Answer: $\frac{ \pm \sqrt{2}+4}{12}$.

## Grading Criteria:

- 15 points - The correct answer is obtained with proper justification.
- 10 points - The solution contains a correct sequence of steps leading to the correct answer, but one arithmetic error is made during the solution.
- 5 points - The solution contains correct steps, at least one answer is obtained with proper justification, but not all cases are considered.
- $\mathbf{0}$ points - The solution does not meet any of the criteria listed.