Solve the equation $15x^3-23x^2+9x=0$ the roots are in H.P

Given $f\left(x\right)=$$15x^3-23x^2+9x-1=0\to \left(1\right)$

The roots are in H.P

$$\begin{gathered} f\left(\frac{1}{x}\right)=\frac{15}{x^3}-\frac{23}{x^2}+\frac{9}{x}-1=0 \\ 15-23x+9x^2-x^3=0 \\ {x}^3-9x^2+23x-15=0\to \left(2\right) \end{gathered}$$

Then roots are in A.P

The roots are $a-d, a, a+d.$

$$\begin{gathered} S_1=a-d+a+a+d=\frac{-\left(-9\right)}{1} \\ \left(\therefore S_1=\frac{-b}{a}\right) \\ 3a=9\Rightarrow a=3 \end{gathered}$$

$$\begin{gathered} \left(x-3\right)\left(x^2-6x+5\right)=0 \\ \left(x-3\right)\left(x^2-5x-x+5\right)=0 \\ \left(x-3\right)\left(\left(x-5\right)x-1\left(x+5\right)\right)=0 \\ \left(x-3\right)\left(x-5\right)\left(x-1\right)=0 \end{gathered}$$

The roots are 1,3, 5

The roots of $f\left(x\right) are 1, \frac{1}{3}, \frac{1}{5}$