sqrt((sqrt(3)-2)^2)+sqrt[3](-8)

To solve the expression \(\sqrt{(\sqrt{3} - 2)^2} + \sqrt{-8}\), we need to handle each part of the expression separately.<br /><br />First, consider \(\sqrt{(\sqrt{3} - 2)^2}\):<br /><br />\[<br />\sqrt{(\sqrt{3} - 2)^2} = |\sqrt{3} - 2|<br />\]<br /><br />Since \(\sqrt{3} \approx 1.732\) is less than 2, \(\sqrt{3} - 2\) is negative. Therefore, the absolute value is:<br /><br />\[<br />|\sqrt{3} - 2| = 2 - \sqrt{3}<br />\]<br /><br />Next, consider \(\sqrt{-8}\):<br /><br />\[<br />\sqrt{-8} = \sqrt{8} \cdot \sqrt{-1} = \sqrt{8} \cdot i = 2\sqrt{2} \cdot i<br />\]<br /><br />Now, combine the results:<br /><br />\[<br />\sqrt{(\sqrt{3} - 2)^2} + \sqrt{-8} = (2 - \sqrt{3}) + 2\sqrt{2}i<br />\]<br /><br />So, the final answer is:<br /><br />\[<br />(2 - \sqrt{3}) + 2\sqrt{2}i<br />\]