There are $6$ pieces of cheese of different weights. For any two pieces we can identify the heavier piece. Given that it is possible to divide them into two groups of equal weights with three pieces in each. Give the explicit way to find these groups by performing two weightings on a regular balance.
To solve the problem of dividing the six pieces of cheese into two groups of equal weight using only two weighings on a regular balance, we can follow these steps:

1. **Define Majorization**:
   - Call a decreasing sequence \(a, b, c\) to *majorize* another decreasing sequence \(d, e, f\) if \(a > d\), \(b > e\), and \(c > f\). This implies that \(a + b + c > d + e + f\), and thus their sums are not equal.

2. **List Possible Groupings**:
   - Suppose the pieces are labeled from heaviest to lightest as \(6, 5, 4, 3, 2, 1\). We need to find a grouping of three pieces each such that their sums are equal. We will consider the group containing the heaviest piece, \(6\), and check all possible combinations:
     - \(6, 5, 4\) majorizes \(3, 2, 1\), so this is impossible.
     - \(6, 5, 3\) majorizes \(4, 2, 1\), so this is impossible.
     - \(6, 5, 2\) majorizes \(4, 3, 1\), so this is impossible.
     - \(6, 5, 1\) does not majorize \(4, 3, 2\). For example, setting \(6 = 100\) kg, \(5 = 99\) kg, \(4 = 98\) kg, \(3 = 97\) kg, \(2 = 85\) kg, \(1 = 81\) kg, we obtain two groups that sum to \(200\) kg each.
     - \(6, 4, 3\) majorizes \(5, 2, 1\), so this is impossible.
     - \(6, 4, 2\) majorizes \(5, 3, 1\), so this is impossible.
     - \(6, 4, 1\) does not majorize \(5, 3, 2\). For example, setting \(6 = 100\) kg, \(5 = 99\) kg, \(4 = 98\) kg, \(3 = 97\) kg, \(2 = 83\) kg, \(1 = 81\) kg, we obtain two groups that sum to \(200\) kg each.
     - \(6, 3, 2\) does not majorize \(5, 4, 1\). For example, setting \(6 = 100\) kg, \(5 = 99\) kg, \(4 = 98\) kg, \(3 = 96\) kg, \(2 = 82\) kg, \(1 = 81\) kg, we obtain two groups that sum to \(200\) kg each.
     - \(6, 3, 1\) does not majorize \(5, 4, 2\). For example, setting \(6 = 100\) kg, \(5 = 98\) kg, \(4 = 97\) kg, \(3 = 96\) kg, \(2 = 82\) kg, \(1 = 81\) kg, we obtain two groups that sum to \(200\) kg each.
     - \(6, 2, 1\) does not majorize \(5, 4, 3\). For example, setting \(6 = 100\) kg, \(5 = 96\) kg, \(4 = 84\) kg, \(3 = 83\) kg, \(2 = 82\) kg, \(1 = 81\) kg, we obtain two groups that sum to \(200\) kg each.

3. **Identify Potential Groupings**:
   - From the above analysis, the possible groupings are:
     - \(6, 5, 1\) vs \(4, 3, 2\)
     - \(6, 4, 1\) vs \(5, 3, 2\)
     - \(6, 3, 1\) vs \(5, 4, 2\)
     - \(6, 2, 1\) vs \(5, 4, 3\)
     - \(6, 3, 2\) vs \(5, 4, 1\)

4. **Perform Two Weighings**:
   - Weigh \(6, 4, 1\) vs \(5, 3, 2\) and \(6, 3, 1\) vs \(5, 4, 2\). If either of these weighings balance, we have found the correct grouping. If not, analyze the results:
     - If \(6, 4, 1\) is heavier and \(6, 3, 1\) is heavier, then the correct grouping is \(6, 2, 1\) vs \(5, 4, 3\).
     - If \(6, 4, 1\) is lighter and \(6, 3, 1\) is lighter, then the correct grouping is \(6, 5, 1\) vs \(4, 3, 2\).
     - If \(6, 4, 1\) is lighter and \(6, 3, 1\) is heavier, this is impossible as moving the \(4\) from the lighter side with the \(3\) shouldn't make the lighter side suddenly heavier.
     - If \(6, 4, 1\) is heavier and \(6, 3, 1\) is lighter, then the correct grouping is \(6, 3, 2\) vs \(5, 4, 1\).

The final answer is \( \boxed{ 6, 5, 1 } \) vs \(4, 3, 2\), \(6, 4, 1\) vs \(5, 3, 2\), \(6, 3, 1\) vs \(5, 4, 2\), \(6, 2, 1\) vs \(5, 4, 3\), or \(6, 3, 2\) vs \(5, 4, 1\).