Find the term independent of `x` in the expansion of `(1+x+2x^3)[(3x^2//2)-(1//3)]^9`

Explanation

(r +1) th term in the expansion of ` ((3)/(2) x^(2) - (1)/(3x))^(9)`
i.e., ` T_(r+ 1) = ""^(9)C_(r) ((3)/(2) ^(2))^(9-r) (-(1)/(3x))^(r)`
` ""^(9)C_(r) ((3)/(2))^(9-r) *x^(18-2r)*(-(1)/(3))^(r) *x^(-r)`
`= ""^(9)C_(r) ((3)/(2))^(9-r)* (-(1)/(3))^(r)*x^(18-3r)`
Here , general term in the expansion of `(1 + x + 2x^(3))`
`((3)/(2) x^(2) -(1)/(3x))^(9) = ""^(9)C_(r) ((3)/(2))^(9-r) (-(1)/(2))^(r) *x^(18-3r)`
` + ""^(9)C_(r) ((3)/(2))^(9-r) (-(1)/(3))^(r) x^(19-3r)`
` +2 ""^(9)C_(r) ((3)/(2))^(9-r) (-(1)/(3))^(r) x^(21-3r)`
For indepandent term putting ` 18-3r = 0,19-3r = 0 `
21 - 3r = 0 respectively , we get
r = 6,r = 19 /3 [impossible ] r= 7 , second term do not given
the indepandent term
Hence , coefficient indepandent of x
` = ""^(9)C_(6) ((3)/(2))^(3)* (-(1)/(3))^(6) + 0 + 2* ""^(9)C_(7)* ((3)/(2))^(2) (-(1)/(3))^(7)`
` = ""^(9)C_(3) * (27)/(8.729) - 2. ""^(9)C_(2) *9/4*1/2187= 7/18-2/27 = 17/54 `