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$(1)$ The ratio 2a, 3b and the duplicate of $9{b^2}:ab$

$(2)$ The sub duplicate ratio of 64: 9, and the ratio 27: 56

$(3)$ The duplicate ratio of $\dfrac{{2a}}{b}:\dfrac{{\sqrt C {a^2}}}{{{b^2}}}$, and the ratio 3ax: 2by

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Hint – In this question we have to deal with terms like duplicate, sub duplicate and we need to take out ratio compounded of ratio and duplicate. Sub duplicate ratio and ratio and duplicate ratio and ratio, so use the basic definition of duplicate, sub-duplicate directly along with the basic formula. Implementation of these two will get you to the answer.

Complete step-by-step answer:

Let us assume two ratios $\left( {x:y} \right){\text{ & }}\left( {p:q} \right)$

Duplicate of the ratio $\left( {p:q} \right)$ is $\left( {{p^2}:{q^2}} \right)$

So, the compound of $\left( {x:y} \right)$ and duplicate of $\left( {p:q} \right)$is $ \Rightarrow \left( {x \times {p^2}} \right):\left( {y \times {q^2}} \right)$

Now as we know that the sub duplicate of the ratio $\left( {p:q} \right)$ is $\left( {\sqrt p :\sqrt q } \right)$

So, the compound of $\left( {x:y} \right)$ and sub duplicate of $\left( {p:q} \right)$ is $ \Rightarrow \left( {x \times \sqrt p } \right):\left( {y \times \sqrt q } \right)$

So, use these properties in the given question we have,

$\left( 1 \right)$ Duplicate of the ratio $\left( {9{b^2}:ab} \right)$ is $\left( {81{b^4}:{a^2}{b^2}} \right)$

So, the compound of $\left( {2a:3b} \right)$ and duplicate of $\left( {9{b^2}:ab} \right)$is $ \Rightarrow \left( {2a \times 81{b^4}} \right):\left( {3b \times {a^2}{b^2}} \right)$

Now simplify the above ratio we have,

$ \Rightarrow \left( {2a \times 81{b^4}} \right):\left( {3b \times {a^2}{b^2}} \right)$

Divide by $3a{b^3}$ we have,

$ = 54b:a$

$\left( 2 \right)$ The sub duplicate of the ratio $\left( {64:9} \right)$ is \[\left( {\sqrt {64} :\sqrt 9 } \right) = \left( {8:3} \right)\]

So, the compound of $\left( {27:56} \right)$ and sub duplicate of $\left( {8:3} \right)$is $ \Rightarrow \left( {27 \times 8} \right):\left( {56 \times 3} \right)$

Now divide by 24 we have

$ \Rightarrow \left( {27 \times 8} \right):\left( {56 \times 3} \right) = 9:7$

$\left( 3 \right)$ Duplicate of the ratio $\left( {\dfrac{{2a}}{b}:\dfrac{{\sqrt c {a^2}}}{{{b^2}}}} \right)$ is $\left( {\dfrac{{4{a^2}}}{{{b^2}}}:\dfrac{{c{a^4}}}{{{b^4}}}} \right)$

So, the compound of $\left( {3ax:2by} \right)$ and duplicate of $\left( {\dfrac{{2a}}{b}:\dfrac{{\sqrt c {a^2}}}{{{b^2}}}} \right)$is $ \Rightarrow \left( {3ax \times \dfrac{{4{a^2}}}{{{b^2}}}} \right):\left( {2by \times \dfrac{{c{a^4}}}{{{b^4}}}} \right)$

Now simplify the above ratio we have,

$ \Rightarrow \left( {3ax \times \dfrac{{4{a^2}}}{{{b^2}}}} \right):\left( {2by \times \dfrac{{c{a^4}}}{{{b^4}}}} \right) = \left( {12x \times \dfrac{{{a^3}}}{{{b^2}}}} \right):\left( {2y \times \dfrac{{c{a^4}}}{{{b^3}}}} \right)$

Divide by $\dfrac{{2{a^3}}}{{{b^2}}}$ we have,

$ = 6x:\dfrac{{yac}}{b} = 6bx:acy$

So, these are the required answers.

Note – Whenever we face such types of problems the key point that we need to have in our mind is that these all are basic definitions along with direct based questions. So having a good understanding of this direct concept helps you solve problems of this kind.

Complete step-by-step answer:

Let us assume two ratios $\left( {x:y} \right){\text{ & }}\left( {p:q} \right)$

Duplicate of the ratio $\left( {p:q} \right)$ is $\left( {{p^2}:{q^2}} \right)$

So, the compound of $\left( {x:y} \right)$ and duplicate of $\left( {p:q} \right)$is $ \Rightarrow \left( {x \times {p^2}} \right):\left( {y \times {q^2}} \right)$

Now as we know that the sub duplicate of the ratio $\left( {p:q} \right)$ is $\left( {\sqrt p :\sqrt q } \right)$

So, the compound of $\left( {x:y} \right)$ and sub duplicate of $\left( {p:q} \right)$ is $ \Rightarrow \left( {x \times \sqrt p } \right):\left( {y \times \sqrt q } \right)$

So, use these properties in the given question we have,

$\left( 1 \right)$ Duplicate of the ratio $\left( {9{b^2}:ab} \right)$ is $\left( {81{b^4}:{a^2}{b^2}} \right)$

So, the compound of $\left( {2a:3b} \right)$ and duplicate of $\left( {9{b^2}:ab} \right)$is $ \Rightarrow \left( {2a \times 81{b^4}} \right):\left( {3b \times {a^2}{b^2}} \right)$

Now simplify the above ratio we have,

$ \Rightarrow \left( {2a \times 81{b^4}} \right):\left( {3b \times {a^2}{b^2}} \right)$

Divide by $3a{b^3}$ we have,

$ = 54b:a$

$\left( 2 \right)$ The sub duplicate of the ratio $\left( {64:9} \right)$ is \[\left( {\sqrt {64} :\sqrt 9 } \right) = \left( {8:3} \right)\]

So, the compound of $\left( {27:56} \right)$ and sub duplicate of $\left( {8:3} \right)$is $ \Rightarrow \left( {27 \times 8} \right):\left( {56 \times 3} \right)$

Now divide by 24 we have

$ \Rightarrow \left( {27 \times 8} \right):\left( {56 \times 3} \right) = 9:7$

$\left( 3 \right)$ Duplicate of the ratio $\left( {\dfrac{{2a}}{b}:\dfrac{{\sqrt c {a^2}}}{{{b^2}}}} \right)$ is $\left( {\dfrac{{4{a^2}}}{{{b^2}}}:\dfrac{{c{a^4}}}{{{b^4}}}} \right)$

So, the compound of $\left( {3ax:2by} \right)$ and duplicate of $\left( {\dfrac{{2a}}{b}:\dfrac{{\sqrt c {a^2}}}{{{b^2}}}} \right)$is $ \Rightarrow \left( {3ax \times \dfrac{{4{a^2}}}{{{b^2}}}} \right):\left( {2by \times \dfrac{{c{a^4}}}{{{b^4}}}} \right)$

Now simplify the above ratio we have,

$ \Rightarrow \left( {3ax \times \dfrac{{4{a^2}}}{{{b^2}}}} \right):\left( {2by \times \dfrac{{c{a^4}}}{{{b^4}}}} \right) = \left( {12x \times \dfrac{{{a^3}}}{{{b^2}}}} \right):\left( {2y \times \dfrac{{c{a^4}}}{{{b^3}}}} \right)$

Divide by $\dfrac{{2{a^3}}}{{{b^2}}}$ we have,

$ = 6x:\dfrac{{yac}}{b} = 6bx:acy$

So, these are the required answers.

Note – Whenever we face such types of problems the key point that we need to have in our mind is that these all are basic definitions along with direct based questions. So having a good understanding of this direct concept helps you solve problems of this kind.