Yes -- easy for Mathematica:

(* the list of equations *)

eqs = {
   d - g == a (d - 1) + b (d - 1)^2 + c (d - 1)^3,
   e - h == a (e - 1) + b (e - 1)^2 + c (e - 1)^3,
   f - j == a (f - 1) + b (f - 1)^2 + c (f - 1)^3
   };

(* the solution as a list of rules *)

sol = Solve[eqs, {a, b, c}]

(*{{a\[Rule]-((-((-1+d)^3 (-1+f)^2-(-1+d)^2 (-1+f)^3) ((-1+e)^3 \
(d-g)-(-1+d)^3 (e-h))+((-1+d)^3 (-1+e)^2-(-1+d)^2 (-1+e)^3) ((-1+f)^3 \
(d-g)-(-1+d)^3 (f-j)))/(((-1+d)^3 (-1+e)^2-(-1+d)^2 (-1+e)^3) \
(-(-1+d)^3 (1-f)+(1-d) (-1+f)^3)-(-(-1+d)^3 (1-e)+(1-d) (-1+e)^3) \
((-1+d)^3 (-1+f)^2-(-1+d)^2 (-1+f)^3))),b\[Rule]-((-3 d^2 e+d^3 e+3 d \
e^2-d e^3+3 d^2 f-d^3 f-3 e^2 f+e^3 f-3 d f^2+3 e f^2+d f^3-e f^3+2 e \
g-3 e^2 g+e^3 g-2 f g+3 e^2 f g-e^3 f g+3 f^2 g-3 e f^2 g-f^3 g+e f^3 \
g-2 d h+3 d^2 h-d^3 h+2 f h-3 d^2 f h+d^3 f h-3 f^2 h+3 d f^2 h+f^3 \
h-d f^3 h+2 d j-3 d^2 j+d^3 j-2 e j+3 d^2 e j-d^3 e j+3 e^2 j-3 d e^2 \
j-e^3 j+d e^3 j)/((-1+d) (d-e) (-1+e) (d-f) (e-f) \
(-1+f))),c\[Rule]-((-d^2 e+d e^2+d^2 f-e^2 f-d f^2+e f^2+e g-e^2 g-f \
g+e^2 f g+f^2 g-e f^2 g-d h+d^2 h+f h-d^2 f h-f^2 h+d f^2 h+d j-d^2 \
j-e j+d^2 e j+e^2 j-d e^2 j)/((-1+d) (d-e) (-1+e) (d-f) (e-f) \
(-1+f)))}}*)

(* your set of values as a list of rules *)

values = {
   d -> 1.8926846,
   e -> 2.5686994,
   f -> 3.3756357,
   g -> 1.892797681,
   h -> 2.568917298,
   j -> 3.376008599
   };

(* substtute the values into the solution *)

sol /. values

(*{{a\[Rule]-0.0001145856497967586`,b\[Rule]-0.000010955938570381746`,\
c\[Rule]-2.8979714654621227`*^-6}} *)

(* If you want a more readable list of the solution execute this *)

{a, b, c} /. sol // Simplify // Transpose // TableForm

I attach the Mathematica file. Note that I used lowercase symbols to avoid conflict with reserved names.

Attachments:

JF.nb