The roots $\alpha , \beta$ and $\gamma$of an equation $x^3-3a{x}^2+3bx-c=0$ are in H.P. Then,

Given that the roots of the equation 

$x^3-3a{x}^2+3bx-c=0$ are in H.P. Therefore, the roots of the reciprocal equation i.e. $c{y}^3-3b{y}^2+3ay-1=0$ are in A.P.

i.e. $\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma }$ are in A.P.

$\frac{2}{\beta }=\frac{1}{\alpha }+\frac{1}{\gamma }$

$\Rightarrow \frac{3}{\beta }=\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\Rightarrow \frac{3}{\beta }=\frac{3b}{c}\Rightarrow \beta =\frac{c}{b} \left[\because \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{3b}{c}\right]$