Question

Radical din 12(radical din 3 - radical din 2 + radical din 6)

Answer 1

$\it \sqrt{12}(\sqrt3-\sqrt2+\sqrt6)=\sqrt{36}-\sqrt{24}+\sqrt{72}=6-\sqrt{4\cdot6}+\sqrt{36\cdot2}=\\ \\ =6-2\sqrt6+6\sqrt2\\ \\ --------------------\\ \\ \sqrt{12}(\sqrt3-\sqrt2+\sqrt6)=\sqrt{4\cdot3}(\sqrt3-\sqrt2+\sqrt6)=2\sqrt3}(\sqrt3-\sqrt2+\sqrt6)=\\ \\ =2(\sqrt9-\sqrt6+\sqrt{18})=2(3-\sqrt6+3\sqrt2)=6-2\sqrt6+6\sqrt2$