A system of bivariate cubics

Question

It is asked to solve for rational numbers $x$ and $y$ the following pair of equations: $$ a=x^3+3xy^2\left(\frac {a-b^3}{3b}\right), $$ $$ \frac {9b^3+a}{3b}=3x^2y+y^3\left(\frac {a-b^3}{3b}\right) $$ Making the following substitutions, $$ \left(\frac {a-b^3}{3b}\right)=\mu, $$ $$ \frac {9b^3+a}{3b}=\kappa $$ The equations are reduced to, $$ a=x^3+3\mu xy^2, $$ $$ \kappa=3x^2y+\mu y^3 $$ By substitution I do get free of one of the variables, but the resultant is a $9^{\text {th}}$ degree equation in $x$, $$ 64b^3x^9-48ab^3x^6-(81b^9-63ab^6-2a^2b^3-a^3)x^3-a^3b^3=0 $$ Trust me or not, but Maxima does solve the equation giving $9$ distinct roots, most of which are imaginary, only one being real, but unfortunately irrational. Thus at some point I do have had some mistake, but I don't know where. Otherwise it would have given at least one simple (rational) root. Can someone please ponit me out the right way? Or is there some more novel method to solve such equations? Any help would be appreciated.