## Task B-2.7.

Let the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by,

$$
f(x)=\frac{2}{\sqrt[3]{x^{2}+4 x+4}+\sqrt[3]{x^{2}-4}+\sqrt[3]{x^{2}-4 x+4}}
$$

What is $f(4)+f(8)+f(12)+\cdots+f(2024)$?
## Solution.

Notice that the expression on the right side can be written as

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

Also, observe that the denominator is of the form $a^{2}+a b+b^{2}$, where $a=\sqrt[3]{x+2}, b=\sqrt[3]{x-2}$. By multiplying the numerator and the denominator by $a-b=\sqrt[3]{x+2}-\sqrt[3]{x-2}$, we will rationalize the denominator of the given fraction. Then we have

$$
\begin{aligned}
& f(x)=\frac{2}{\sqrt[3]{(x+2)^{2}}+\sqrt[3]{(x+2)(x-2)}+\sqrt[3]{(x-2)^{2}}} \cdot \frac{\sqrt[3]{x+2}-\sqrt[3]{x-2}}{\sqrt[3]{x+2}-\sqrt[3]{x-2}}= \\
& =\frac{2(\sqrt[3]{x+2}-\sqrt[3]{x-2})}{\sqrt[3]{(x+2)^{3}}-\sqrt[3]{(x-2)^{3}}}= \\
& =\frac{2(\sqrt[3]{x+2}-\sqrt[3]{x-2})}{(x+2)-(x-2)}=\frac{2(\sqrt[3]{x+2}-\sqrt[3]{x-2})}{4}=\frac{\sqrt[3]{x+2}-\sqrt[3]{x-2}}{2}
\end{aligned}
$$

It follows that

$$
\begin{aligned}
& f(4)+f(8)+f(12)+\cdots+f(2024)= \\
& =\frac{1}{2}(\sqrt[3]{6}-\sqrt[3]{2}+\sqrt[3]{10}-\sqrt[3]{6}+\sqrt[3]{14}-\sqrt[3]{10}+\cdots+\sqrt[3]{2026}-\sqrt[3]{2022})= \\
& =\frac{1}{2}(\sqrt[3]{2026}-\sqrt[3]{2})
\end{aligned}
$$

## SCHOOL COMPETITION IN MATHEMATICS

3rd grade - high school - B variant
January 4, 2024.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.