\(x^a\cdot x^b\,=\,x^{a+b}\)
So....
\(\sqrt3^\sqrt2\cdot\sqrt3^{-\sqrt2}\,=\,\sqrt3^{\sqrt2+-\sqrt2}\\~\\ \phantom{\sqrt3^\sqrt2\cdot\sqrt3^{-\sqrt2}}\,=\,\sqrt3^{\sqrt2-\sqrt2}\\~\\ \phantom{\sqrt3^\sqrt2\cdot\sqrt3^{-\sqrt2}}\,=\,\sqrt3^0\\~\\ \phantom{\sqrt3^\sqrt2\cdot\sqrt3^{-\sqrt2}}\,=\,1\)
