## Task A-2.4. (4 points)

For which $x \in \mathbb{R}$ is the number $\sqrt[3]{4+4 x}$ greater than the number $1+\sqrt[3]{x}$?
## Solution.

We seek all $x \in \mathbb{R}$ that satisfy the inequality $\sqrt[3]{4+4 x}>1+\sqrt[3]{x}$. It is equivalent to the following inequalities:

$$
\begin{gathered}
\sqrt[3]{4+4 x}>1+\sqrt[3]{x} \\
4+4 x>1+3 \sqrt[3]{x}+3 \sqrt[3]{x^{2}}+x \\
x-\sqrt[3]{x^{2}}-\sqrt[3]{x}+1>0 \\
\left(\sqrt[3]{x^{2}}-1\right)(\sqrt[3]{x}-1)>0 \\
(\sqrt[3]{x}-1)^{2}(\sqrt[3]{x}+1)>0
\end{gathered}
$$

The first factor is strictly positive for all $x \neq 1$, so the entire expression on the left side is positive when $x \neq 1$ and $\sqrt[3]{x}+1>0$, i.e., for $x \in\langle-1,1\rangle \cup\langle 1,+\infty\rangle$.

The number $\sqrt[3]{4+4 x}$ is greater than the number $1+\sqrt[3]{x}$ if and only if $x$ is greater than -1 and different from 1.

Note: The problem can also be solved by analyzing the factors in $\left({ }^{*}\right)$.