a) \({{\sqrt a - \sqrt b } \over {\root 4 \of a - \root 4 \of b }} - {{\sqrt a + \root 4 \of {ab} } \over {\root 4 \of a + \root 4 \of b }} = {{\left( {\root 4 \of a + \root 4 \of b } \right)\left( {\root 4 \of a - \root 4 \of b } \right)} \over {\root 4 \of a - \root 4 \of b }} - {{\root 4 \of a \left( {\root 4 \of a + \root 4 \of b } \right)} \over {\root 4 \of a + \root 4 \of b }}\)
\( = \root 4 \of a + \root 4 \of b - \root 4 \of a = \root 4 \of b \)
b) \({{a - b} \over {\root 3 \of a - \root 3 \of b }} - {{a + b} \over {\root 3 \of a + \root 3 \of b }} = {{{{\left( {\root 3 \of a } \right)}^3} - {{\left( {\root 3 \of b } \right)}^3}} \over {\root 3 \of a - \root 3 \of b }} - {{{{\left( {\root 3 \of a } \right)}^3} + {{\left( {\root 3 \of b } \right)}^3}} \over {\root 3 \of a + \root 3 \of b }}\)
\( = \root 3 \of {{a^2}} + \root 3 \of {ab} + \root 3 \of {{b^2}} - \left( {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} } \right) = 2\root 3 \of {ab} \)
c) \(\left( {{{a + b} \over {\root 3 \of a + \root 3 \of b }} - \root 3 \of {ab} } \right):{\left( {\root 3 \of a - \root 3 \of b } \right)^2} = \left( {\root 3 \of {{a^2}} - \root 3 \of {ab} + \root 3 \of {{b^2}} - \root 3 \of {ab} } \right):{\left( {\root 3 \of a - \root 3 \of b } \right)^2}\)
\( = \left( {\root 3 \of {{a^2}} - 2\root 3 \of {ab} + \root 3 \of {{b^2}} } \right):{\left( {\root 3 \of a - \root 3 \of b } \right)^2} = {\left( {\root 3 \of a - \root 3 \of b } \right)^2}:{\left( {\root 3 \of a - \root 3 \of b } \right)^2} = 1\)
d) \({{a - 1} \over {{a^{{3 \over 4}}} + {a^{{1 \over 2}}}}}.{{\sqrt a + \root 4 \of a } \over {\sqrt a + 1}}.{a^{{1 \over 4}}} + 1. = {{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)} \over {\sqrt a \left( {\root 4 \of a + 1} \right)}}.{{\root 4 \of a \left( {\root 4 \of a + 1} \right)} \over {\left( {\sqrt a + 1} \right)}}.\root 4 \of a + 1\)
\( = \sqrt a - 1 + 1 = \sqrt a \).
