(i) Let the mean proportional between
\[6\text{ }+\text{ }3\surd 3~and\text{ }8\text{ }-\text{ }4\surd 3~\]
be x.
So,
\[6\text{ }+\text{ }3\surd 3,\text{ }x\text{ }and\text{ }8\text{ }-\text{ }4\surd 3\]
are in continued proportion.
\[\begin{array}{*{35}{l}}
6\text{ }+\text{ }3\surd 3~:\text{ }x\text{ }=\text{ }x\text{ }:\text{ }8\text{ }-\text{ }4\surd 3 \\
x~\times ~x\text{ }=\text{ }\left( 6\text{ }+\text{ }3\surd 3 \right)\text{ }\left( 8\text{ }-\text{ }4\surd 3 \right) \\
{{x}^{2~}}=\text{ }48\text{ }+\text{ }24\surd 3\text{ }-\text{ }24\surd 3~\text{ }-36 \\
{{x}^{2~}}=\text{ }12 \\
x=\text{ }2\surd 3 \\
\end{array}\]
(ii) Let the mean proportional between a – b and a3 – a2b be x.
a – b, x, a3 – a2b are in continued proportion.
\[\begin{array}{*{35}{l}}
a\text{ }-\text{ }b:\text{ }x\text{ }=\text{ }x:\text{ }{{a}^{3}}~-\text{ }{{a}^{2}}b \\
x\text{ }x\text{ }x\text{ }=\text{ }\left( a\text{ }-\text{ }b \right)\text{ }\left( {{a}^{3}}~-\text{ }{{a}^{2}}b \right) \\
{{x}^{2}}~=\text{ }\left( a\text{ }-\text{ }b \right)\text{ }{{a}^{2}}\left( a\text{ }-\text{ }b \right)\text{ }=\text{ }{{\left[ a\left( a\text{ }-\text{ }b \right) \right]}^{2}} \\
x\text{ }=\text{ }a\left( a\text{ }-\text{ }b \right) \\
\end{array}\]