Answer
The 6th term is $\binom{8}{5}3^3(2)^5x^3=56\times 27\times 32x^3=48384x^3$
Work Step by Step
By using the binomial theorem, we can expand the algebraic expression in the form of:
$(ax+b)^n=\binom{n}{0}a^nb^0x^n+\binom{n}{1}a^{n-1}b^1x^{n-1}+...+\binom{n}{n-i}a^{i}b^{n-i}x^{i}+...+\binom{n}{n-1}a^1b^{n-1}x^1+\binom{n}{n}a^0b^nx^0$
We expand the given expression:
$(3x+2)^8=\binom{8}{0}3^8(2)^0x^8+\binom{8}{1}3^7(2)^1x^7+\binom{8}{2}3^6(2)^2x^6+\binom{8}{3}3^5(2)^3x^5+\binom{8}{4}3^4(2)^4x^4+\binom{8}{5}3^3(2)^5x^3+\binom{8}{6}3^2(2)^6x^2+\binom{8}{7}3^1(2)^7x^1+\binom{8}{8}3^0(2)^8x^0$
The 6th term is $\binom{8}{5}3^3(2)^5x^3=56\times 27\times 32x^3=48384x^3$
