Find the missing number 38, 1368, 54758, 89, 7743, 704702, 67, 4355, ?

Actually the Answer is '300562'
Most likely a typo
Explanation:-
If we take the series as: a, b, c
then the 3rd term i.e. c = axb +2b +a
Here c = 67x4355 + 2x4355 + 67
= 291785 + 8710 + 67
= 300562

38*38-38*2=1368; 38*38*38-38*3=54758 similarly 67*67*67-67*3

38,38*36,[38*(38-1)+(38-3)]*38=38, 1368, 54758(First three)
89,89*87,[89*(89-1)+(89-3)]*89 =89, 7743, 704702 (Next three numbers),
67,67*65,[67*(67-1)+(67-3)]*67= 67,4335,300562 (It was given as 300568
Hence my choice is for 8.

logic
split in 3 groups (38,1368,54758),(89,7743,704702)),(67,4355,...)
second number of 1st group :-
38 * ( 38 - 2)= 1368
3rd Number of 1st group
{
add ((38-2)*2 + 1) =73 to 2nd number
i.e. 1368 + 73 = 1441
}
so 3rd number is
38 * 1441 = 54758

similarly in 3rd group
2nd number:-
67 * 65 = 4355
3rd number should be
((65*2 + 1) + 4355) * 67
=( 131+ 4355) *67
= 4486 * 67 = 300562 ...( I guess 8th option is misprint for 300568 instead of 300562)

or if I am wrong then please admin don't remove my logic from submission, Because it is also a valid reason, but correct if it is wrong.

(38,1368,54758),(89,7743,704702),(67,4355,____)
2nd term (1368) = (38^2) - (38*2) = 1444 - 76 = 1368
3rd term (54758) = (38^3) - (38*3) = 54872 - 114 = 54758
Similarly,
2nd term = 4355 = (67^2) - (67*2) = 4489 - 134 = 4355
3rd term = (67^3) - (67*3) = 300763 - 201 = 300562

38,(38×38-38×2),(38×38×38-38×3),89,(89×89-89×2),(89×89×89-89×3),67,(67×67-67×2),(67×67×67-67×3)
(67×67×67-67×3)=300562
not match any one of them. given above answer
So correct answer is 300562

( 38 * 38 ) - ( 2 * 38 ) = 1368
( 1368 * 38 ) + ( 2 * 1368 ) + 38 = 54758

similarly,
( 67 * 67 ) - ( 2 * 67 ) = 4355
( 67 * 4355 ) + ( 2 * 4355 ) + 67 = 300562

67
67^2 - (67*2)=4355
4355* (67+2) +67 = 300562

the answer should be 300562 but since the nearest option was option { 8 } so i take it as the printing mistake
the reason is as follows:

the pattern we observe here is
first group the sequence in 3 elements each so that

{ 38 , 1368 , 54758 } { 89 , 7743 , 704702 } { 67 , 4355 , ? }

the first no. in each group is the key to obtain other no.(s) in that group

let first no. be x then ,
second no. = x * ( x - 2 )
and third no. = x * { second no. + ( 2 * ( x-2 ) ) + 1 }
By applying the same rule,
for first group ,
first no. = 38
second no. = 38 * ( 38 - 2 ) = 1368
third no. = 38 * { 1368 + ( 2 * ( 38 - 2 ) ) + 1 }
= 54758
for second group
first no. = 89
second no. = 89 * ( 89 - 2 ) = 7743
third no. = 89 * { 7743 + ( 2 * ( 89 - 2 )) + 1}
= 704702
similarly for the third group
first no. = 67
second no. = 67 * ( 67 - 2 ) = 4355
third no. = 67 * { 4355 + ( 2 * ( 67 - 2)) + 1}
= 300562 answer
rounded answer according to options
300568

take group of 3 consecutive no.
and apply like this : a, a(a-2), a(a^3 - 3)
where 'a' is first no. of three consecutive.

e.g. take this 38, 1368, 54758
here "a" = 38
so, 38, 38(38-2), 38(38^2-3) and so on

But my point of view answer should be 300562 because when we take a=67 and apply on above formulae we get this number.