Question

If the roots of the cubic equation $a{x}^3+b{x}^2+cx+d=0$ are in G.P., then

• A. $c^{3}a=b^{3}d$
• B. $c{a}^3=b{d}^3$
• C. $a^{3}b=c^{3}d$
• D. $a{b}^3=c{d}^3$

Answer 1

The correct option is A

$c^{3}a=b^{3}d$

Step-1: Determine the sum and product of roots of the given equation $a{x}^3+b{x}^2+cx+d=0$

Let $\alpha ,\beta ,and\gamma$are the roots of the given equation

$$\begin{gathered} \alpha +\beta +\gamma =\frac{-b}{a}...\left(1\right) \\ \alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}...\left(2\right) \\ \alpha \beta \gamma =\frac{-d}{a}...\left(3\right) \end{gathered}$$

Since $\alpha ,\beta ,and\gamma$are in G.P.

Let $r$ be the common ratio, the roots are as

$$\begin{gathered} \alpha =\alpha \\ \beta =\alpha r \\ \gamma =\alpha r^2 \end{gathered}$$

Substituting the above in equations (1), (2), and (3).

$$\begin{gathered} \left(1\right)\Rightarrow \alpha +\alpha r+\alpha r^2=\frac{-b}{a} \\ \alpha (1+r+r^2)=\frac{-b}{a}......\left(4\right) \\ \left(2\right)\Rightarrow \alpha \left(\alpha r\right)+\left(\alpha r\right)\left(\alpha r^2\right)+\left(\alpha r^2\right)\left(\alpha \right)=\frac{c}{a} \\ {\alpha }^{2}r+{\alpha }^2{r}^3+{\alpha }^2{r}^2=\frac{c}{a} \\ {\alpha }^{2}r(1+r+r^2)=\frac{c}{a}...\left(5\right) \\ \left(3\right)\Rightarrow \alpha \left(\alpha r\right)\left(\alpha r^2\right)=\frac{-d}{a} \\ \Rightarrow {\alpha }^3{r}^3=\frac{-d}{a}...\left(6\right) \end{gathered}$$

Step-2: Solve equations (4), (5), and (6) for a, b, and c.

Divide equation (4) by (5)

$$\begin{gathered} \frac{\alpha (1+r+r^2)}{\alpha r(1+r+r^2)}=\frac{{\displaystyle \frac{-b}{a}}}{{\displaystyle \frac{c}{a}}} \\ \frac{1}{\alpha r}=\frac{-b}{a}*\frac{a}{c} \\ \frac{1}{\alpha r}=\frac{-b}{c} \\ \alpha r=\frac{-c}{b} \end{gathered}$$

Step-3: Put $\alpha r=\frac{-c}{b}$in equation (6)

$$\begin{gathered} {\left(\frac{-c}{b}\right)}^3=\frac{-d}{a} \\ \Rightarrow -\frac{c^3}{b^3}=\frac{-d}{a} \\ Dividebothsidesby-1 \\ \frac{c^3}{b^3}=\frac{d}{a} \\ \Rightarrow a{c}^3=d{b}^3 \\ \Rightarrow c^{3}a=b^{3}d \end{gathered}$$

Therefore, option (1) is the correct option.