Let the width of the strips cut off be \( x \) cm. After cutting the strips, the new dimensions of the rectangle will be:
- Length: \( 17 - x \) cm
- Width: \( 14 - x \) cm
The area \( A \) of the remaining rectangle is given as 108 cm². Thus, we can set up the equation:
\[ (17 - x)(14 - x) = 108 \]
Expanding this equation, we have:
\[ 238 - 17x - 14x + x^2 = 108 \]
This simplifies to:
\[ x^2 - 31x + 238 - 108 = 0 \]
\[ x^2 - 31x + 130 = 0 \]
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 \( c = 130 \). Plugging in these values:
\[ x = \frac{31 \pm \sqrt{(-31)^2 - 4 \cdot 1 \cdot 130}}{2 \cdot 1} \] \[ x = \frac{31 \pm \sqrt{961 - 520}}{2} \] \[ x = \frac{31 \pm \sqrt{441}}{2} \] \[ x = \frac{31 \pm 21}{2} \]
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 \( x = 26 \) is not a valid solution (as it exceeds both dimensions of the rectangle), we have:
\[ x = 5 \]
Thus, the width of the strips removed is \( \boxed{5} \) cm.