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**Hint**: At first, take roots of the equation as $\alpha ,\alpha r,\alpha {{r}^{2}}$ with r as common ratio, then, use the relation between the roots and coefficient to get relation between a, b, c, d. Sum of roots is related to the ratio of –b and a of the quadratic equation whereas the product of roots is given by ratio of -d and a. Then we have another relation which gives the sum of product of two consecutive roots as ratio of c and a.

__Complete step-by-step answer____:__

In the question, if the roots of cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d\] are in G.P, then, we have to find a relation between a, b, c, d.

Now, as we know that, the roots are in G.P. So, let's suppose the roots are \[\alpha ,\alpha r,\alpha {{r}^{2}}\] with a common ratio 'r'.

So, by relation between the roots and the coefficient of equation, we can say that,

Sum of the roots \[\Rightarrow -\dfrac{b}{a}\]

\[\Rightarrow \alpha +\alpha r+\alpha {{r}^{2}}=-\dfrac{b}{a}\]

Which can be written as,

\[\alpha \left( 1+r+{{r}^{2}} \right)=-\dfrac{b}{a}\]

Product of the roots \[\Rightarrow \dfrac{c}{a}\]

\[\begin{align}

& \Rightarrow \alpha \times \alpha r+\alpha \times \alpha {{r}^{2}}+\alpha r\times \alpha {{r}^{2}}=\dfrac{c}{a} \\

& \Rightarrow {{\alpha }^{2}}r+{{\alpha }^{2}}{{r}^{2}}+{{\alpha }^{2}}{{r}^{3}}=\dfrac{c}{a} \\

\end{align}\]

Which can be written as,

\[{{\alpha }^{2}}r\left( 1+r+{{r}^{2}} \right)=\dfrac{c}{a}\]

We can written it as,

\[\alpha r\times \alpha \left( 1+r+{{r}^{2}} \right)=\dfrac{c}{a}\]

Now, we can write \[\alpha \left( 1+r+{{r}^{2}} \right)\text{ as -}\dfrac{b}{a}\] as we know that \[\alpha \left( 1+r+{{r}^{2}} \right)\text{ as -}\dfrac{b}{a}\]

\[\begin{align}

& \alpha r\times -\dfrac{b}{a}=\dfrac{c}{a} \\

& \Rightarrow \alpha r=-\dfrac{c}{b} \\

\end{align}\]

The final relation between roots and coefficient of equation is,

\[\begin{align}

& \alpha \times \alpha r\times \alpha {{r}^{2}}=-\dfrac{d}{a} \\

& \Rightarrow {{\alpha }^{3}}{{r}^{3}}=-\dfrac{d}{a} \\

\end{align}\]

We know that, \[\alpha r=-\dfrac{c}{b}\]

So, we can write,

\[{{\alpha }^{3}}{{r}^{3}}={{\left( \alpha r \right)}^{3}}\Rightarrow {{\left( -\dfrac{c}{b} \right)}^{3}}\]

We found that \[{{\alpha }^{3}}{{r}^{3}}\text{ is }-\dfrac{d}{a}\] so, we can write,

\[\begin{align}

& {{\left( -\dfrac{c}{b} \right)}^{3}}=-\dfrac{d}{a} \\

& \Rightarrow -\dfrac{{{c}^{3}}}{{{b}^{3}}}=-\dfrac{d}{a} \\

\end{align}\]

Now, on cross multiplication we get,

\[{{c}^{3}}a={{b}^{3}}d\]

**So, the correct answer is “Option A”.**

**Note**: When it is said that, roots are given in G.P, then we can take the value of three roots as any three consecutive terms of any G.P such as \(\alpha ,\alpha r,\alpha {r^2}\) by reducing the three variables into two. Also, while taking this, one should know here 'r' which is a common ratio cannot be equal to '0'. If 'r' is equal to '0' then, it will not satisfy the condition of being in G.P.

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