(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}\]