Show that (sqrt(3) + sqrt(75))^{2} = 108

The key here is to simplify the left hand side. There are two different approaches to take here, one slightly faster but both perfectly legitimate. First approach: Remember the formula (a + b)^{2} = a^{2} + 2ab + b^{2}. Then (sqrt(3) + sqrt(75))^{2} = 3 + 2sqrt(3)sqrt(75) + 75 = 78 + 2sqrt(225) = 78 + 2*15 = 108. Second approach: This approach is effectively the same as the first but in slightly more steps (which should be easier in general). We can write the left hand side out in full as (sqrt(3) + sqrt(75)) (sqrt(3) + sqrt(75)). From here, recall how we multiply these kinds of brackets together: (a + b)(c + d) = ac + ad + bc + bd. So we have sqrt(3)*sqrt(3) + sqrt(3)*sqrt(75) + sqrt(75)*sqrt(3) + sqrt(75)sqrt(75) = 3 + sqrt(225) + sqrt(225) + 75 = 78 + 215 = 108.