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Hint: Try to find the relation between a, b, c using geometric progression definition and then try to use it in the series of ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ and prove that it is in GP.

In the question we are given that a, b, c are in G.P.

The above statement means that there exists a number ‘r’ such that ‘a’ when multiplied with ‘r’ gives ‘b’ and ‘b’ when multiplied with ‘r’ gives ‘c’.

Then we can represent this above relation as,

$b=ar$ and $c=br$

So, $c=br=\left( ar \right)\times r=a{{r}^{2}}$

Hence, the numbers $a,b,c$ can be represented as $a,ar,a{{r}^{2}}$.

Now, we are asked about ${{a}^{3}},{{b}^{3}},{{c}^{3}}$.

So, we can transform ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ as ${{a}^{3}},{{a}^{3}}{{r}^{3}},{{a}^{3}}{{r}^{6}}$.

Here, in the ${{a}^{3}},{{a}^{3}}{{r}^{3}},{{a}^{3}}{{r}^{6}}$ we see that there exists a new number ${{r}^{3}}$ which multiplied to ${{a}^{3}}$ become ${{a}^{3}}{{r}^{3}}$ which is ${{b}^{3}}$, which when further multiplied becomes ${{a}^{3}}{{r}^{6}}$ which is ${{c}^{3}}$.

Geometric progression series is a series in which the ratio between two consecutive series is the same.

So we will check here, by substituting the values,

$\dfrac{{{b}^{3}}}{{{a}^{3}}}=\dfrac{{{(ar)}^{3}}}{{{a}^{3}}}$

Cancelling the like terms, we get the ratio as

$\dfrac{{{b}^{3}}}{{{a}^{3}}}={{r}^{3}}.........(i)$

Similarly,

$\dfrac{{{c}^{3}}}{{{b}^{3}}}=\dfrac{{{(a{{r}^{2}})}^{3}}}{{{\left( ar \right)}^{3}}}$

Cancelling the like terms, we get the ratio as

$\dfrac{{{c}^{3}}}{{{b}^{3}}}={{r}^{3}}.........(ii)$

Comparing equations (i) and (ii), we see that the ratios are the same.

Hence, ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ are in $G.P.$ is proved.

Note: There is a shortcut to the above question, if a, b, c are in G.P, there is a formula which is ${{b}^{2}}=ac$. Now for proving that ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ is also in $G.P.,$ we can simply substitute $'a'$ by $'{{a}^{3}}'$, $'b'$ by $'{{b}^{3}}'$, $'c'$ by $'{{c}^{3}}'$. Hence, we get ${{b}^{6}}={{a}^{3}}{{c}^{3}}$ which satisfy the fact that ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ is in $G.P.$

In the question we are given that a, b, c are in G.P.

The above statement means that there exists a number ‘r’ such that ‘a’ when multiplied with ‘r’ gives ‘b’ and ‘b’ when multiplied with ‘r’ gives ‘c’.

Then we can represent this above relation as,

$b=ar$ and $c=br$

So, $c=br=\left( ar \right)\times r=a{{r}^{2}}$

Hence, the numbers $a,b,c$ can be represented as $a,ar,a{{r}^{2}}$.

Now, we are asked about ${{a}^{3}},{{b}^{3}},{{c}^{3}}$.

So, we can transform ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ as ${{a}^{3}},{{a}^{3}}{{r}^{3}},{{a}^{3}}{{r}^{6}}$.

Here, in the ${{a}^{3}},{{a}^{3}}{{r}^{3}},{{a}^{3}}{{r}^{6}}$ we see that there exists a new number ${{r}^{3}}$ which multiplied to ${{a}^{3}}$ become ${{a}^{3}}{{r}^{3}}$ which is ${{b}^{3}}$, which when further multiplied becomes ${{a}^{3}}{{r}^{6}}$ which is ${{c}^{3}}$.

Geometric progression series is a series in which the ratio between two consecutive series is the same.

So we will check here, by substituting the values,

$\dfrac{{{b}^{3}}}{{{a}^{3}}}=\dfrac{{{(ar)}^{3}}}{{{a}^{3}}}$

Cancelling the like terms, we get the ratio as

$\dfrac{{{b}^{3}}}{{{a}^{3}}}={{r}^{3}}.........(i)$

Similarly,

$\dfrac{{{c}^{3}}}{{{b}^{3}}}=\dfrac{{{(a{{r}^{2}})}^{3}}}{{{\left( ar \right)}^{3}}}$

Cancelling the like terms, we get the ratio as

$\dfrac{{{c}^{3}}}{{{b}^{3}}}={{r}^{3}}.........(ii)$

Comparing equations (i) and (ii), we see that the ratios are the same.

Hence, ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ are in $G.P.$ is proved.

Note: There is a shortcut to the above question, if a, b, c are in G.P, there is a formula which is ${{b}^{2}}=ac$. Now for proving that ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ is also in $G.P.,$ we can simply substitute $'a'$ by $'{{a}^{3}}'$, $'b'$ by $'{{b}^{3}}'$, $'c'$ by $'{{c}^{3}}'$. Hence, we get ${{b}^{6}}={{a}^{3}}{{c}^{3}}$ which satisfy the fact that ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ is in $G.P.$

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