After some algebra on the roots, I get $ab(b-2a^2)=18$ which seems to give only solution $(a,b)=(2,9)$.

For the first equation the sum of the roots is $3a$ and they are in AP,hence one of the roots is $a$.Hence we get the relation $ab+18c=2a^3$.Similarly $c$ is a root of the second equation,hence we get the relation $bc^2+c=0$,i.e $c=0$ or $bc=-1$.Now $c=0$ implies all the roots of the second equation are zero since if one term in a GP is zero then all the terms are zero,but that is impossible since the GP is non-constant.Hence $bc=-1$.Putting this value in the first relation we get $b^2 a-2a^3 b-18=0$. Hence $b=a^2\pm\sqrt{a^4+\frac{18}{a}}$.Hence $a=1,2,3,6,9,18$.Out of these six values only $a=2$ gives a positive integral value of $b=9$.We can check that for these values of $a,b$ the conditions are satisfied.

can this be done using discriminant of a cubic equation

Equation1:- \[ x^3-3ax^2+bx+18c=0 \]Roots form a non constant A.P. Let the roots be $\alpha$,$\beta$,$\gamma$.As the roots are in A.P. 2$\beta$=$\alpha$+$\gamma$.. ->$\alpha$= $\beta-d$,$\beta$=$\beta$,$\gamma$=$\beta+d$(where d is the common difference) $\sigma_1$ $\longrightarrow$ 3$\beta$=3$a$$\longrightarrow$ $\beta$=$a$ $\sigma_2$ $\longrightarrow$ $a^2 - ad +a^2 +ad +a^2 - d^2$=$b$$\longrightarrow$ 3$a^2-d^2$=$b$ $\sigma_3$ $\longrightarrow$ $a(a^2-d^2)$=$-18c$ As $\beta$=$a$ is a root, satisfying it in the equation gives $ab-2a^3=-18c$ Equation 2\[ x^3+bx^2+x-c^3=0 \]Let the roots be $u$,$v$,$w$.then $u$=$\frac{v}{r}$,$v$=$v$,$w$=$vr$(where $r$ is the common ratio). $\sigma_1$$\longrightarrow$$v(\frac{1}{r} + 1 + r)$=-$b$ $\sigma_2$$\longrightarrow$$v^2(\frac{1}{r} + 1 + r)$=$1$ $\sigma_1$ and $\sigma_2$ imply that $\frac{1}{v}$=$-b$ $\sigma_3$$\longrightarrow$ $v^3=c^3$$\longrightarrow$ $v=c$ as $\frac{1}{v}$=$-b$ and $v=c$ we have $bc=-1$ As $ab-2a^3=-18c$ and $bc=-1$ we have $ab(b-2a^2)$=$18$and the positive integral roots are $(a,b)=(2,9)$. we are done.