If the roots of $k{x}^3-18x^2-36x+8=0$ are in H.P., then k =

The given equation is $k{x}^3-18x^2-36x+8=0$

given that roots are in A.P.

now consider $f\left(\frac{1}{x}\right)=0$

$\Rightarrow k{\left(\frac{1}{x}\right)}^3-18{\left(\frac{1}{x}\right)}^2-36\left(\frac{1}{x}\right)+8=0$

$\Rightarrow 8x^3-36x^2-18x+k=0$ …. ( 1 )

Now the roots are in A.P.

i.e., $a-d,a,a+d$  

$\therefore$ sum of the roots : $a-d+a+a+d=\frac{36}{8}$

$3a=\frac{36}{8}\Rightarrow a=\frac{3}{2}$

$\therefore$ $a$  is one root of the equation Eq. ( 1 )

$f\left(\frac{3}{2}\right)=0$

$\Rightarrow 8{\left(\frac{3}{2}\right)}^3-36{\left(\frac{3}{2}\right)}^2-18\left(\frac{3}{2}\right)+k=0$

$\Rightarrow k=81$