This is an extension of the things you learned in \emph{Topic 1: Algebra}. We will be using 3 main results and then replacing the $a$'s and $b$'s with \emph{surds}. \begin{align} & (a+b)^{2}=a^{2}+2 a b+b^{2} \\ & (a-b)^{2}=a^{2}-2 a b+b^{2} \\ & (a-b)(a+b)=a^{2}-b^{2} \end{align} \begin{examplebox} \subsection{Examples:} \begin{questions} \Question[5] \begin{doublespace} \begin{parts} \part \[(\sqrt{5}+\sqrt{3})^{2}=\] \begin{solutionordottedlines}[2cm] $(\sqrt{5}+\sqrt{3})^{2} = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{3}) + (\sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}$ \end{solutionordottedlines} \part \[(3 \sqrt{2}-2 \sqrt{3})^{2}=\] \begin{solutionordottedlines}[2cm] $(3 \sqrt{2}-2 \sqrt{3})^{2} = (3\sqrt{2})^2 - 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 9\cdot2 - 2\cdot6\sqrt{6} + 4\cdot3 = 18 - 12\sqrt{6} + 12$ \end{solutionordottedlines} \part \[(2 \sqrt{3}+4 \sqrt{6})^{2}=\] \begin{solutionordottedlines}[2cm] $(2 \sqrt{3}+4 \sqrt{6})^{2} = (2\sqrt{3})^2 + 2(2\sqrt{3})(4\sqrt{6}) + (4\sqrt{6})^2 = 4\cdot3 + 2\cdot8\sqrt{18} + 16\cdot6 = 12 + 16\sqrt{18} + 96$ \end{solutionordottedlines} \part \[(\sqrt{7}-5)(\sqrt{7}+5)=\] \begin{solutionordottedlines}[2cm] $(\sqrt{7}-5)(\sqrt{7}+5) = (\sqrt{7})^2 - (5)^2 = 7 - 25 = -18$ \end{solutionordottedlines} \part \[(5 \sqrt{6}-2 \sqrt{5})(5 \sqrt{6}+2 \sqrt{5})=\] \begin{solutionordottedlines}[2cm] $(5 \sqrt{6}-2 \sqrt{5})(5 \sqrt{6}+2 \sqrt{5}) = (5\sqrt{6})^2 - (2\sqrt{5})^2 = 25\cdot6 - 4\cdot5 = 150 - 20 = 130$ \end{solutionordottedlines} \end{parts} \end{doublespace} \end{questions} \end{examplebox}