Question

If the roots of 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 $c^{3}a=b^{3}d$

$a{x}^3+b{x}^2+cx+d=0$

Let the three roots be

$\alpha =\frac{\lambda }{r}$

$\beta =\lambda$

$\gamma =\lambda r$

$\alpha +\beta +\gamma =-\frac{b}{a}$

$\frac{\lambda }{r}+\lambda +\lambda r=-\frac{b}{a}$

$\lambda (\frac{1}{r}+1+r)=-\frac{b}{a}$ ................1

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$

$\frac{\lambda }{r}\times \lambda +\lambda \times \lambda r+\lambda r\times \frac{\lambda }{r}=\frac{c}{a}$

${\lambda }^2(\frac{1}{r}+1+r)=\frac{c}{a}$ ................2

$\alpha \beta \gamma =-\frac{d}{a}$

$\frac{\lambda }{r}\times \lambda \times \lambda r=-\frac{d}{a}$

${\lambda }^3=-\frac{d}{a}$ ..................3

dividing eq.2 by 1

$\frac{{\lambda }^2(\frac{1}{r}+1+r)}{\lambda (\frac{1}{r}+1+r)}=\frac{\frac{c}{a}}{\frac{-b}{a}}=-\frac{c}{b}$

$\lambda =-\frac{c}{b}$

${\lambda }^3=-\frac{c^3}{b^3}$

from eq.3

$-\frac{d}{a}=-\frac{c^3}{b^3}$

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