Answer:
third differences are all 6a
Step-by-step explanation:
Substituting for x, we find the sequence values to be ...
(a+b+c+d), (8a+4b+2c+d), (27a+9b+3c+d), (64a+16b+4c+d), (125a+25b+5c+d),
(216a+36b+6c+d)
Then first differences are ...
(7a+3b+c), (19a+5b+c), (37a+7b+c), (61a+9b+c), (91a+11b+c)
Second differences are ...
(12a+2b), (18a+2b), (24a+2b), (30a+2b)
And the third differences are ...
(6a), (6a), (6a) . . . . . constant
_____
Additional comment
These results can help you write a polynomial relation up to 3rd degree for a sequence that has constant differences at any level up to 3rd differences.
Of course, we find the differences by subtracting each term from the one following.
