Note: This is a classic Example of Self Dual function.
For any logical Expression, two times dual gives the same expression.
In Self dual Expression one time dual gives the same expression.
$f = (A+B).(B+C).(C+A) $
$f^d = AB + BC + CA$
The dual does not look like the original function but simple boolean simplification leads to the original function.
$f^d = AB + CA + BC$
CA = AC (Commutative)
$f^d = AB + AC + BC$
now taking A common
$f^d = A (B + C) + BC$
now OR is distributive over AND
$f^d = (A + BC)(B + C + BC)$
here B + BC = B (absorption law)
A + BC = (A + B)(A + C) (+ is distributive over . )
$f^d = (A + B)(A + C)(B + C)$
now after applying commutativity, we get
$f^d = (A+B).(B+C).(C+A) $
Therefore
$f^d = f$
The special thing about the given boolean expression is that it is a self dual function.