Let the width of the strips cut off from one side and one end be \( x \) cm.
After cutting off the strips, the dimensions of the remaining piece are:
- Length: \( 17 - x \) cm
- Width: \( 14 - x \) cm
The area of the remaining piece is given by:
\[ (17 - x)(14 - x) = 108 \]
Now, let's expand this equation:
\[ 238 - 17x - 14x + x^2 = 108 \]
Simplifying it, we obtain:
\[ x^2 - 31x + 238 - 108 = 0 \]
\[ x^2 - 31x + 130 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
Here, \( a = 1 \), \( b = -31 \), and \( c = 130 \).
Calculating the discriminant:
\[ b^2 - 4ac = (-31)^2 - 4 \cdot 1 \cdot 130 = 961 - 520 = 441 \]
Now substituting back into the quadratic formula:
\[ x = \frac{{31 \pm \sqrt{441}}}{2} = \frac{{31 \pm 21}}{2} \]
Calculating the two possible values for \( x \):
- \( x = \frac{{31 + 21}}{2} = \frac{52}{2} = 26 \)
- \( x = \frac{{31 - 21}}{2} = \frac{10}{2} = 5 \)
Since \( x = 26 \) cm is not possible (as it exceeds the small side of the cardboard), we have:
\[ x = 5 \text{ cm} \]
Thus, the width of the strips removed is \( \boxed{5} \) cm.