Question

if the roots of the cubic equation ax3+bx2+cx+d=0 are in g.p then
(a)c³a=b³d
(b)a³b=c³d

Answer 1

$$\begin{gathered} a{x}^3+b{x}^2+cx+d=0 \\ Letthethreerootsbe \\ \alpha =\frac{\lambda }{r} \\ \beta =\lambda \\ \gamma =\lambda r \\ \alpha +\beta +\gamma =-\frac{b}{a} \\ \frac{\lambda }{r}+\lambda +\lambda r=-\frac{b}{a} \\ \lambda \left(\frac{1}{r}+r+1\right)=-\frac{b}{a}\_\_\_\_\_\_\left(1\right) \\ \alpha \beta +\beta \gamma +\alpha \gamma =\frac{c}{a} \\ \frac{\lambda }{r}\times \lambda +\lambda \times \lambda r+\lambda r\times \frac{\lambda }{r}=\frac{c}{a} \\ \frac{{\lambda }^2}{r}+{\lambda }^{2}r+{\lambda }^2=\frac{c}{a} \\ {\lambda }^2\left(\frac{1}{r}+r+1\right)=\frac{c}{a}\_\_\_\_\_\_\left(2\right) \\ \alpha \beta \gamma =-\frac{d}{a}\_\_\_\_\_\_\_\_\_\_\left(3\right) \\ \frac{\lambda }{r}\times \lambda \times \lambda r=-\frac{d}{a} \\ {\lambda }^3=-\frac{d}{a}\_\_\_\_\_\_\_\_\left(1\right) \\ Dividingequation2by1 \\ \frac{{\lambda }^2\left(\frac{1}{r}+r+1\right)}{\lambda \left(\frac{1}{r}+r+1\right)}=\frac{\frac{c}{a}}{-{\displaystyle \frac{b}{a}}}=-\frac{c}{b} \\ \lambda =-\frac{c}{b}\_\_\_\_\_\_\_\_\_\left(2\right) \\ {\lambda }^3=-\frac{c^3}{b^3} \\ fromequation3 \\ -\frac{c^3}{b^3}=-\frac{d}{a} \\ {c}^{3}a=b^{3}d \end{gathered}$$