The condition that the roots of $x^3-b x^2+c x-d=0$ are in geometric progression is

Equation, $x^3-b x^2+c x-d=0$ Let the roots of this cubic equation in GP are $\left(\frac{a}{r}, a, a r\right)$ Then, Sum of the roots $\frac{a}{r}+a+a r=-\left(\frac{-b}{1}\right) \Rightarrow b$ $\Rightarrow \quad a\left(\frac{1}{r}+1+r\right)=b$ $\ldots$ (i) and $\quad \frac{a}{r} \cdot a+a \cdot a r+a r \cdot \frac{a}{r}=\frac{c}{1}$ $\Rightarrow \quad a^2\left(\frac{1}{r}+r+1\right)=c$ $\ldots$ (ii) Product of the roots $\frac{a}{r} \cdot a \cdot a r=-\left(\frac{-d}{1}\right)=d$ $\Rightarrow \quad a^3=d$ $\ldots$ (iii) Eq. (ii) divided by Eq. (i), we get $a=c / b$, put the value of ' $a$ ' in Eq. (iii) $(c / b)^3=d$ $\Rightarrow \quad c^3=b^3 d$