Let's determine the distinct digits $A, B, C, D, E, F$ such that the following equalities hold:

$$
\begin{aligned}
& A B C^{2}=D A E C F B \quad \text { and } \\
& C B A^{2}=E D C A B F
\end{aligned}
$$
I. solution. From the two equations written, we see that the square of both three-digit numbers is six-digit, so the value of $A$ and $C$ as the leading digits of the bases must be at least 3. According to the condition, different letters represent different digits, and the last digit of the square of $A$ and $C$ as the bases ends in $B$ and $F$, respectively, so neither $A$ nor $C$ can be 5 or 6. Therefore, the possible values of $A$ and $C$ are

$$
3,4,7,8,9
$$

the endings of their squares are respectively

$$
9,6,9,4,1
$$

these give the possible values of $B$ and $F$.

The last two digits of a square number are not affected by the digit in the hundreds place of the base, so in the first equation, we can examine the relationship between $BC$ and $FB$ without knowing the value of $A$.

From the first equation, we see that the value of $B$ is given by the last digit of the square of $C$, for example, if $C=7$, then $B=9, BC=97$. The value of $FB$ can be calculated by squaring: $FB=09$; then $BF=90$, which, according to the second equation, is the two-digit ending of the square of $BA$. However, there is no square number ending in 90, so this cannot be a solution to the problem.

Let's take the possible values of $BC$ in order. According to (3) and (4), these are 93, 64, 97 (which we have already examined), 48, 19. The endings of their squares are: $49,96,09,04,61$. The reverse order of the remaining $94,69,40$ and 16 is that 94 and 40 are not square endings, because if the last digit of a square number is 0, then 0 also stands before it, and if the last digit is 4, then an even digit stands before it. In the two remaining cases, $BF=69$ or $BF=16$.

If $BF=69$, then $A=3$ or $A=7$, and in the first case

$$
\begin{array}{ll}
ABC=364, & ABC^{2}=132496 \\
CBA=463, & CBA^{2}=214369
\end{array}
$$

satisfy the conditions.

If, however, $A=7$, then $ABC=764, CBA=467$. From the first equation, the last two digits of $BC^{2}$ are $FB=96, B=6$, and from the second equation, the last two digits of $BA^{2}$ are $BE=89, B=8$. This is a contradiction for $B$.

In the other case, $ABC=419, CBA=914$. Similarly to the previous, the last two digits of $BA^{2}$ are $BF=96$, and the last two digits of $BC^{2}$ are $FB=61$, which is again a contradiction for $B$.

In summary, the problem has only one solution, as we have seen: $A=3, B=6, C=4, D=1, E=2$ and $F=9$.

II. solution. The two-digit ending of the two squares is the reverse of each other. Based on this, we can narrow down the number of endings to be examined. For this purpose, we show that after the ending of $25^{2}$, no new two-digit ending of square numbers is found, the previously found ones are repeated in reverse order, and after $50^{2}$, the segment between $1^{2}$ and $50^{2}$ is repeated.

Indeed, if $x$ is an integer and $0 \leq x \leq 25$, then $50 \geq 50-x \geq 25$, and the two-digit ending of $(50-x)^{2}$ is the same as that of $x^{2}$, because their difference is:

$$
(50-x)^{2}-x^{2}=50(50-2 x)=100(25-x)
$$

on the other hand

$$
(50+x)^{2}-x^{2}=50(50+2 x)=100(25+x)
$$

both are divisible by 100, and this proves our statements.

Among the 22 different endings up to $25^{2}$, the two-digit endings formed from different digits and their reverses are 16 and 61, as well as 69 and 96. These first appear in the squares of 4 and 19, and 13 and 14, respectively, so according to the previous

$\ldots 16$ ends with the ... $04, \ldots 46, \ldots 54, \ldots 96$ numbers squared

$\ldots 61$ ends with the ... $19, \ldots 31, \ldots 69, \ldots 81$ numbers squared

$\ldots 69$ ends with the ... $13, \ldots 37, \ldots 63, \ldots 87$ numbers squared

$\ldots 96$ ends with the ... $14, \ldots 36, \ldots 64, \ldots 86$ numbers squared.

Among the endings of the bases found this way, more can be ruled out based on the following two observations. In both (1) and (2), the ending of the base is different from that of its square, the tens digit of the base is repeated in the two-digit ending of its square, in (1) in the second digit of the ending, and in (2) in the first digit of the ending.

In (5), we marked with bold digits the endings of the bases that correspond to these, the ones marked with ordinary digits cannot provide a solution. And since the ... 16 ending of the square has become impossible, the ... 61 ending is also missing, so - if there is a solution at all - it can only be the completion of the following:

(1) from: $\ldots 64^{2}=\ldots 96$

(2) from: $\ldots 63^{2}=\ldots 69$,

so $B=6, C=4, A=3, F=9$, thus the two base numbers are 364 and 463.

Their squares are respectively: 132496,214369, which also meet the requirements for the remaining digits, $D=1, E=2$. We have completed the solution.