While a simple closed-form solution isn't available, you could try series expansions in the parameter $d$, if $d$ is small.  Thus suppose you want the root that is $a$ at $d=0$.  For simplicity, rewrite the equation as
$$ X (X-B)(X-C) = d(X-E)(X-F)$$
where $X=x-a$, $B = b-a$, $C=c-a$, $E=e-a$, $F=f-a$.  Then this root is
$$\eqalign{ X &= {\frac {EF}{BC}}d+ \left( -{\frac {{E}^{2}F}{{B}^{2}{C}^{2}}}-{\frac 
{E{F}^{2}}{{B}^{2}{C}^{2}}}+{\frac {{E}^{2}{F}^{2}}{{B}^{2}{C}^{3}}}+{
\frac {{E}^{2}{F}^{2}}{{B}^{3}{C}^{2}}} \right) {d}^{2}\cr+ &\left( {
\frac {{E}^{3}F}{{B}^{3}{C}^{3}}}+3\,{\frac {{E}^{2}{F}^{2}}{{B}^{3}{C
}^{3}}}+{\frac {E{F}^{3}}{{B}^{3}{C}^{3}}}-3\,{\frac {{E}^{3}{F}^{2}}{
{B}^{3}{C}^{4}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{3}{C}^{4}}}+2\,{
\frac {{E}^{3}{F}^{3}}{{B}^{3}{C}^{5}}}-3\,{\frac {{E}^{3}{F}^{2}}{{B}
^{4}{C}^{3}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{4}{C}^{3}}}+3\,{\frac {{
E}^{3}{F}^{3}}{{B}^{4}{C}^{4}}}+2\,{\frac {{E}^{3}{F}^{3}}{{B}^{5}{C}^
{3}}} \right) {d}^{3}\cr
\cr+&\ldots}
$$
This (and as many terms as you want) can be obtained from the Lagrange–Bürmann formula.