Ap EapcetBinomial TheoremAP EAPCET 2024 - 20th May Evening ShiftMCQ+1 / -02024If $|x|<\frac{2}{3}$, then the 4th term in the expansion of $(3 x-2)^{\frac{2}{3}}$ is :A$\frac{\sqrt[3]{4}}{6} x^3$B$-\frac{\sqrt[3]{4}}{6} x^3$C$\frac{\sqrt[3]{4}}{8} x^3$D$-\frac{\sqrt[3]{4}}{8} x^3$Check AnswerClear SelectionReveal AnswerShow Explanation
