Sides of a triangle form an arithmetic sequence with common difference $2$, and its area is $6 \text{ cm }^2$. Find its
sides.
1. Let the sides of the triangle be \( x-2, x, x+2 \). Since the sides form an arithmetic sequence with a common difference of 2, this assumption is valid.
2. The semi-perimeter \( s \) of the triangle is given by:
   \[
   s = \frac{(x-2) + x + (x+2)}{2} = \frac{3x}{2}
   \]
3. Using Heron's formula, the area \( A \) of the triangle is:
   \[
   A = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   where \( a = x-2 \), \( b = x \), and \( c = x+2 \). Substituting these values, we get:
   \[
   A = \sqrt{\left(\frac{3x}{2}\right) \left(\frac{3x}{2} - (x-2)\right) \left(\frac{3x}{2} - x\right) \left(\frac{3x}{2} - (x+2)\right)}
   \]
4. Simplifying the terms inside the square root:
   \[
   \frac{3x}{2} - (x-2) = \frac{3x}{2} - x + 2 = \frac{x}{2} + 2
   \]
   \[
   \frac{3x}{2} - x = \frac{x}{2}
   \]
   \[
   \frac{3x}{2} - (x+2) = \frac{3x}{2} - x - 2 = \frac{x}{2} - 2
   \]
5. Substituting these back into the formula for the area:
   \[
   A = \sqrt{\left(\frac{3x}{2}\right) \left(\frac{x}{2} + 2\right) \left(\frac{x}{2}\right) \left(\frac{x}{2} - 2\right)}
   \]
6. Given that the area \( A \) is \( 6 \text{ cm}^2 \), we have:
   \[
   6 = \sqrt{\left(\frac{3x}{2}\right) \left(\frac{x}{2} + 2\right) \left(\frac{x}{2}\right) \left(\frac{x}{2} - 2\right)}
   \]
7. Simplifying inside the square root:
   \[
   6 = \sqrt{\left(\frac{3x}{2}\right) \left(\frac{x}{2}\right) \left(\frac{x}{2} + 2\right) \left(\frac{x}{2} - 2\right)}
   \]
   \[
   6 = \sqrt{\frac{3x}{2} \cdot \frac{x}{2} \cdot \left(\frac{x^2}{4} - 4\right)}
   \]
   \[
   6 = \sqrt{\frac{3x^2}{4} \cdot \left(\frac{x^2 - 16}{4}\right)}
   \]
   \[
   6 = \sqrt{\frac{3x^2 (x^2 - 16)}{16}}
   \]
8. Squaring both sides to eliminate the square root:
   \[
   36 = \frac{3x^2 (x^2 - 16)}{16}
   \]
9. Multiplying both sides by 16:
   \[
   576 = 3x^2 (x^2 - 16)
   \]
10. Dividing both sides by 3:
    \[
    192 = x^2 (x^2 - 16)
    \]
11. Let \( y = x^2 \). Then the equation becomes:
    \[
    y^2 - 16y - 192 = 0
    \]
12. Solving this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
    \[
    y = \frac{16 \pm \sqrt{16^2 + 4 \cdot 192}}{2}
    \]
    \[
    y = \frac{16 \pm \sqrt{256 + 768}}{2}
    \]
    \[
    y = \frac{16 \pm \sqrt{1024}}{2}
    \]
    \[
    y = \frac{16 \pm 32}{2}
    \]
    \[
    y = 24 \quad \text{or} \quad y = -8
    \]
13. Since \( y = x^2 \) and \( y \) must be positive, we have:
    \[
    y = 24
    \]
    \[
    x^2 = 24
    \]
    \[
    x = 2\sqrt{6}
    \]
14. Therefore, the sides of the triangle are:
    \[
    2\sqrt{6} - 2, \quad 2\sqrt{6}, \quad 2\sqrt{6} + 2
    \]

The final answer is \( \boxed{ 2\sqrt{6} - 2, 2\sqrt{6}, 2\sqrt{6} + 2 } \).