15. A right circular cylinder has a volume of \( (6+2 \sqrt{3}) \pi \mathrm{cm}^{3} \) and a base radius of \( (1+\sqrt{3}) \mathrm{cm} \). Find its height in the form \( (a+b \sqrt{3}) \mathrm{cm} \), where \( a \) and \( b \) are integers.

To find the height of the cylinder, we start with the formula for the volume of a right circular cylinder, which is given by: \[ V = \pi r^2 h \] where \( V \) is the volume, \( r \) is the base radius, and \( h \) is the height. We are provided with the volume \( V = (6 + 2\sqrt{3})\pi \) cm\(^3\) and the base radius \( r = (1 + \sqrt{3}) \) cm. First, we can simplify the relationship to isolate the height: \[ (6 + 2\sqrt{3})\pi = \pi (1 + \sqrt{3})^2 h \] Next, we evaluate \( (1 + \sqrt{3})^2 \): \[ (1 + \sqrt{3})^2 = 1^2 + 2(1)(\sqrt{3}) + (\sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3} \] Now, substituting this back into our volume equation gives us: \[ (6 + 2\sqrt{3})\pi = \pi (4 + 2\sqrt{3}) h \] We can divide both sides by \( \pi \) (assuming \( \pi \neq 0 \)): \[ 6 + 2\sqrt{3} = (4 + 2\sqrt{3}) h \] Next, to find \( h \), we rearrange the equation: \[ h = \frac{6 + 2\sqrt{3}}{4 + 2\sqrt{3}} \] To simplify this, we can multiply the numerator and denominator by the conjugate of the denominator, which is \( 4 - 2\sqrt{3} \): \[ h = \frac{(6 + 2\sqrt{3})(4 - 2\sqrt{3})}{(4 + 2\sqrt{3})(4 - 2\sqrt{3})} \] Calculating the denominator first: \[ (4 + 2\sqrt{3})(4 - 2\sqrt{3}) = 4^2 - (2\sqrt{3})^2 = 16 - 12 = 4 \] Now calculating the numerator: \[ (6 + 2\sqrt{3})(4 - 2\sqrt{3}) = 6 \cdot 4 - 6 \cdot 2\sqrt{3} + 2\sqrt{3} \cdot 4 - 2\sqrt{3} \cdot 2\sqrt{3} \] This simplifies to: \[ 24 - 12\sqrt{3} + 8\sqrt{3} - 12 = 12 - 4\sqrt{3} \] Putting it all together, we have: \[ h = \frac{12 - 4\sqrt{3}}{4} = 3 - \sqrt{3} \] Thus, we can express the height \( h \) in the required form \( (a + b\sqrt{3}) \) cm: \[ h = 3 - 1\sqrt{3} \] where \( a = 3 \) and \( b = -1 \). Therefore, the height of the cylinder is: \[ \boxed{(3 - \sqrt{3})\, \mathrm{cm}} \]