Question

\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)

Answer 1

\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}=-\left(\sqrt{6}+\sqrt{2}\right)\sqrt{\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)}=-\sqrt{2}.\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}=-\left(\sqrt{3}+1\right).\sqrt{2}.\sqrt{\left(\sqrt{3}-2\right).\left(-1\right)}=-\left(\sqrt{3}+1\right).\sqrt{2.\left(2-\sqrt{3}\right)}=-\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=-\left(\sqrt{3}+1\right)\sqrt{3-2\sqrt{3}+1}=-\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}=-\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=-\left(3-1\right)=-2\)