Solution:

We take the hint. This statement is actually the "Radical Axis Theorem." To prove it, we let $O_{1}$, $O_{2}$, $O_{3}$ be the centers of the circles and $r_{1}$, $r_{2}$, $r_{3}$ their respective radii. We shall consider the set of points $P$ satisfying $P O_{1}^{2} - P O_{2}^{2} = r_{1}^{2} - r_{2}^{2}$. For any $P$, let $Q$ be the foot of the perpendicular from $P$ to $O_{1} O_{2}$, and use directed distances on line $O_{1} O_{2}$ (i.e. distances may be positive or negative according to the order of the points). Then, $P O_{1}^{2} - P O_{2}^{2}$ equals
$$\left(P Q^{2} + Q O_{1}^{2}\right) - \left(P Q^{2} + Q O_{2}^{2}\right) = Q O_{1}^{2} - Q O_{2}^{2} = \left(Q O_{1} - Q O_{2}\right)\left(Q O_{1} + Q O_{2}\right) = O_{2} O_{1}\left(O_{2} O_{1} + 2 Q O_{2}\right),$$
a linear function of $Q$; hence there is exactly one position of $Q$ for which $P O_{1}^{2} - P O_{2}^{2} = r_{1}^{2} - r_{2}^{2}$, and the desired set of $P$ is a line perpendicular to $O_{1} O_{2}$ and passing through this $Q$. But, since $A O_{1} = r_{1}$, $A O_{2} = r_{2}$, we see that $A$ lies on this line; similarly, so does $D$. So the line is line $A D$; i.e. $P O_{1}^{2} - P O_{2}^{2} = r_{1}^{2} - r_{2}^{2}$ iff $P$ is on line $A D$. Likewise, one sees that $P O_{1}^{2} - P O_{3}^{2} = r_{1}^{2} - r_{3}^{2}$ iff $P$ is on line $B E$, and $P O_{2}^{2} - P O_{3}^{2} = r_{2}^{2} - r_{3}^{2}$ iff $P$ is on line $C F$. But then, if we take $P$ to be the intersection of $A D$ and $C F$, we have $P O_{1}^{2} - P O_{3}^{2} = \left(P O_{1}^{2} - P O_{2}^{2}\right) + \left(P O_{2}^{2} - P O_{3}^{2}\right) = \left(r_{1}^{2} - r_{2}^{2}\right) + \left(r_{2}^{2} - r_{3}^{2}\right) = r_{1}^{2} - r_{3}^{2}$, so $P$ lies on $B E$. Thus, $A D$, $B E$, $C F$ are concurrent.

Now the quadrilateral $A B D E$ is cyclic with diagonals meeting at $O$, so triangles $A B O$, $E D O$ are similar and $A B / D E = B O / O D$. Similarly, using two other quadrilaterals, we find that $C D / F A = D O / O F$, $E F / B C = F O / O B$. Multiplying together gives
$$\frac{A B \cdot C D \cdot E F}{D E \cdot F A \cdot B C} = \frac{B O}{D O} \cdot \frac{D O}{F O} \cdot \frac{F O}{B O} = 1,$$
which is what we need.