We know about an arithmetic sequence of different numbers that its ninth term is equal to the cube of the second term, and the square and the fourth power of the second term are also terms of the sequence. Let's write down the first two terms of the sequence.
Let the $n$-th term of the sequence be denoted by $a_{n}$. From the properties of the arithmetic sequence, it follows that

$$
a_{n}=a_{2}+(n-2) d \text { if } n \geqq 1
$$

According to the first condition, $a_{9}=a_{2}^{3}$, that is,

$$
a_{2}^{3}=a_{2}+7 d
$$

According to the second and third conditions, there exist positive integers $n$ and $m$ such that $a_{2}^{2}=a_{n}$ and $a_{2}^{4}=a_{m}$, that is,

$$
\begin{aligned}
& a_{2}^{2}=a_{2}+(n-2) d \\
& a_{2}^{4}=a_{2}+(m-2) d
\end{aligned}
$$

Let us introduce the variables $n^{\prime}=n-2$ and $m^{\prime^{\prime}}=m-2$. Then $n^{\prime}, m^{\prime} \geqq-1$ are integers and $n^{\prime}$ is not 0, because in that case $a_{2}=a_{2}^{2}=a_{2}^{3}$, that is, $a_{9}=a_{2}$, which is not possible.

$$
\begin{gathered}
a_{2}^{2}=a_{2}+n^{\prime} d \\
a_{2}^{4}=a_{2}+m^{\prime} d
\end{gathered}
$$

Subtracting $a_{2}$ from both sides of equations (1), (2), and (3), we get

$$
\begin{aligned}
& a_{2}^{3}-a_{2}=a_{2}\left(a_{2}+1\right)\left(a_{2}-1\right)=7 d \\
& a_{2}^{2}-a_{2}=a_{2}\left(a_{2}-1\right)=n^{\prime} d \\
& a_{2}^{4}-a_{2}=a_{2}\left(a_{2}-1\right)\left(a_{2}^{2}+a_{2}+1\right)=m^{\prime} d
\end{aligned}
$$

Dividing the upper and lower equations by the middle one (with $n^{\prime} \neq 0$ and $d \neq 0$),

$$
\begin{gathered}
a_{2}+1=\frac{7}{n^{\prime}} \\
a_{2}^{2}+a_{2}+1=\frac{m^{\prime}}{n^{\prime}}
\end{gathered}
$$

The left side of equation (5) can be written in the form $\left(a_{2}+1\right)^{2}-\left(a_{2}+1\right)+1$, and substituting (4) into this, we get

$$
\frac{m^{\prime}}{n^{\prime}}=\left(\frac{7}{n^{\prime}}\right)^{2}-\frac{7}{n^{\prime}}+1
$$

Multiplying through by $n^{\prime}$ and rearranging, we get $\frac{49}{n^{\prime}}=m^{\prime}+7-n^{\prime}$, which is an integer, so $n^{\prime}$ is a divisor of 49.

Moreover, $n^{\prime} \geqq-1$, so the possible values of $n^{\prime}$ are: $n^{\prime}=-1,1,7,49$.

Here, if $n^{\prime}=-1$, then $m^{\prime}<-1$, which is impossible. If $n^{\prime}=7$, then $m^{\prime}=7$ and thus $a_{2}^{4}=a_{2}^{2}$, which gives $a_{2}=0,1$, or -1. From (1), in all three cases $d=0$, so the terms of the sequence are not distinct. This contradicts the condition of the problem.

The following two possibilities remain:

$$
\begin{aligned}
n^{\prime}=1, & m^{\prime}=43 \\
n^{\prime}=49, & m^{\prime}=43
\end{aligned}
$$

From (4), $a_{2}$ can be calculated, and from (2), $d$ can be calculated: $a_{2}=\frac{7}{n^{\prime}}-1 ; \quad d=\frac{a_{2}^{2}-a_{2}}{n^{\prime}}$.

Thus, two sequences satisfy the conditions of the problem, with their first two terms being:

$$
-24,6 ; \text { or }-\frac{2136}{2401} ;-\frac{6}{7}
$$