Th last four terms of the binomial expansion are 14175x^9y^8 , 20412x^6y^10 , 10206x^3y^12 and 2187y^14.
The formula of binomial expansion is
(x + y)^n = Σ (nCk) x^(n-k) y^k (1)
We have bee given the expression as (2x^3 + 3y^2)^7
Here, x = 2x^3 , y = 3y^2 and n = 7
Putting the required values in equation (1) we get
(2x^3 + 3y^2)^7 = 7C0(2x^3)^7(3y^2)^0 + 7C1(2x^3)^6(3y^2)^1 + 7C2(2x^3)^5(3y^2)^2 + 7C3(2x^3)^4(3y^2)^3 + 7C4(2x^3)^3(3y^2)^4 + 7C5(2x^3)^2(3y^2)^5 + 7C6(2x^3)^1(3y^2)^6 + 7C7(2x^3)^0(3y^2)^7
(2x^3 + 3y^2)^7 = 128x^21 + 7*(64x^18)*3y^2 + 21*(32x^15)*(9y^4) + 35*(16x^12)*(21y^6) + 35*(5x^9)*(81y^8) + 21*(4x^6)*(243y^10) + 7*(2x^3)*(729y^12) + 2187y^14
(2x^3 + 3y^2)^7 = 128x^21 + 1344x^18y^2 + 6048x^15y^4 + 11760x^12y^6 + 14175x^9y^8 + 20412x^6y^10 + 10206x^3y^12 + 2187y^14.
Hence the last four terms of binomial expansion are 14175x^9y^8 , 20412x^6y^10 , 10206x^3y^12 and 2187y^14.
Learn more about binomial expansion here : brainly.com/question/13800206
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