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\[\begin{align}

& A.\text{ }{{c}^{3}}a={{b}^{3}}d \\

& B.\text{ }{{a}^{2}}c={{b}^{2}}d \\

& C.\text{ }a{{c}^{2}}=b{{d}^{2}} \\

& D.\text{ NOTA} \\

\end{align}\]

Answer

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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,

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