The condition that the roots of $x^3-b x^2+c x-d=0$ are in geometric progression is
The condition that the roots of $x^3-b x^2+c x-d=0$ are in geometric progression is
$c^3=b^3 d$
$c^2=b^2 d$
$c=b d^3$
$c=b d^2$
Solution
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)
$(c / b)^3=d$
$\Rightarrow \quad c^3=b^3 d$