Answer
The sixth term is $48384 \ x^3$.
Work Step by Step
According to the binomial theorem, we can expand the algebraic expression in the form of:
$(a+b)^n=\binom{n}{0}b^0a^n+\binom{n}{1} b^1a^{n-1}++\binom{n}{n-i} b^{n-i}a^{i}++\binom{n}{n-1}b^{n-1}a^1+\binom{n}{n}b^nx^0$
Now, we will expand the given expression by replacing $(a+b)^n$ with $(3x+2)^8$.
$(3x+2)^8=\binom{8}{0}(2)^0(3x)^8+\binom{8}{1}(3x)^7(2)^1+\binom{8}{2}(3x)^6(2)^2+\binom{8}{3}(3x)^5(2)^3+\binom{8}{4}(3x)^4(2)^4+\binom{8}{5}(3x)^3(2)^5+\binom{8}{6}(3x)^2(2)^6+\binom{8}{7}(3x)^1(2)^7+\binom{8}{8}(3x)^0(2)^8$
So, the 6th term is $\dbinom{8}{5}3^3(2)^5 \ x^3=56\times 27\times 32 \ x^3=48384 \ x^3$.
