(b) Let $4 + \sqrt {15} = x$, then $4 – \sqrt {15} = {1 \over x}$

$6 + \sqrt {35} = y$, then $6 – \sqrt {35} = {1 \over y}$

$\therefore $ Given expression = ${{{x^{3/2}} + {1 \over {{x^{3/2}}}}} \over {{y^{3/2}} – {1 \over {{y^{3/2}}}}}} = {{{x^3} + 1} \over {{y^3} – 1}}.{\left( {{y \over x}} \right)^{3/2}}$

$ = {{{{(4 + \sqrt {15} )}^3} + 1} \over {{{(6 + \sqrt {35} )}^3} – 1}}\,.\,{\left( {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} \right)^{3/2}}$

$ = {{(4 + \sqrt {15} + 1)\,\{ {{(4 + \sqrt {15} )}^2} – (4 + \sqrt {15} ) + 1\} } \over {(6 + \sqrt {35} – 1)\,\{ {{(6 + \sqrt {35} )}^2} + (6 + \sqrt {35} ) + 1\} }} \times {\left( {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} \right)^{3/2}}$

$ = {{5 + \sqrt {15} } \over {5 + \sqrt {35} }}.\,{{\{ 31 + 8\sqrt {15} – 4 – \sqrt {15} + 1\} } \over {\{ 71 + 12\sqrt {35} + 6 + \sqrt {35} + 1\} }}\,.\,{\left( {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} \right)^{3/2}}$

$ = {{\sqrt 5 + \sqrt 3 } \over {\sqrt 5 + \sqrt 7 }} \times \,{{28 + 7\sqrt {15} } \over {78 + 13\sqrt {35} }}\,{\left( {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} \right)^{3/2}}$

$ = {{\sqrt 5 + \sqrt 3 } \over {\sqrt 5 + \sqrt 7 }}.{7 \over {13}}.\sqrt {{{6 + \sqrt {35} } \over {4 + \sqrt {15} }}} $

$ = {7 \over {13}}\,.\,{{\sqrt 3 + \sqrt 5 } \over {\sqrt 5 + \sqrt 7 }}\,.\,\sqrt {{{{{(\sqrt 5 + \sqrt 7 )}^2}} \over 2}\,.\,{2 \over {{{(\sqrt 3 + \sqrt 5 )}^2}}}} $

$ = {7 \over {13}}\,.\,{{\sqrt 3 + \sqrt 5 } \over {\sqrt 5 + \sqrt 7 }}\,.\,{{\sqrt 5 + \sqrt 7 } \over {\sqrt 3 + \sqrt 5 }} = {7 \over {13}}$.