Answer
$$3i$$
Work Step by Step
$$\eqalign{
& \frac{{ - 3\sqrt 2 + 3i\sqrt 6 }}{{\sqrt 6 + i\sqrt 2 }} \cr
& {\text{Convert the numerator and the denominator to trigonometric form}} \cr
& - 3\sqrt 2 + 3i\sqrt 6 \cr
& r = \sqrt {{{\left( { - 3\sqrt 2 } \right)}^2} + {{\left( {3\sqrt 2 } \right)}^2}} = 6\sqrt 2 \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{3\sqrt 6 }}{{ - 3\sqrt 2 }}} \right) = - {60^ \circ } \cr
& \theta = - {60^ \circ } + {180^ \circ } = {120^ \circ } \cr
& - 3\sqrt 2 + 3i\sqrt 6 = 6\sqrt 2 \left( {\cos \left( {{{120}^ \circ }} \right) + i\sin \left( {{{120}^ \circ }} \right)} \right) \cr
& \cr
& \sqrt 6 + i\sqrt 2 \cr
& r = \sqrt {{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} = 2\sqrt 2 \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{\sqrt 2 }}{{\sqrt 6 }}} \right) = {30^ \circ } \cr
& \sqrt 6 + i\sqrt 2 = 2\sqrt 2 \left( {\cos \left( {{{30}^ \circ }} \right) + i\sin \left( {{{30}^ \circ }} \right)} \right) \cr
& \cr
& {\text{Then,}} \cr
& \frac{{ - 3\sqrt 2 + 3i\sqrt 6 }}{{\sqrt 6 + i\sqrt 2 }} = \frac{{6\sqrt 2 \left( {\cos \left( {{{120}^ \circ }} \right) + i\sin \left( {{{120}^ \circ }} \right)} \right)}}{{2\sqrt 2 \left( {\cos \left( {{{30}^ \circ }} \right) + i\sin \left( {{{30}^ \circ }} \right)} \right)}} \cr
& {\text{Using the Quotient Theorem}} \cr
& = 3\left[ {\cos \left( {{{120}^ \circ } - {{30}^ \circ }} \right) + i\sin \left( {{{120}^ \circ } - {{30}^ \circ }} \right)} \right] \cr
& {\text{Simplify}} \cr
& = 3\left[ {\cos \left( {{{90}^ \circ }} \right) + i\sin \left( {{{90}^ \circ }} \right)} \right] \cr
& {\text{Write in Rectangular form}} \cr
& = 3\left( {0 + i} \right) \cr
& = 3i \cr} $$
