If |x|

Question

(B) Given the expression $(3x - 2)^{2/3}$. For the binomial expansion,we rewrite it as: $(3x - 2)^{2/3} = [-2(1 - \frac{3x}{2})]^{2/3} = (-2)^{2/3} (1

Vedclass · Accepted Answer

$-\frac{\sqrt[3]{4}}{6} x^3$

Vedclass · Answer

(B) Given the expression $(3x - 2)^{2/3}$. For the binomial expansion,we rewrite it as: $(3x - 2)^{2/3} = [-2(1 - \frac{3x}{2})]^{2/3} = (-2)^{2/3} (1 - \frac{3x}{2})^{2/3} = \sqrt[3]{4} (1 - \frac{3x}{2})^{2/3}$. The general term $T_{r+1}$ in the expansion of $(1+y)^n$ is $\binom{n}{r} y^r$. Here $n = \frac{2}{3}$ and $y = -\frac{3x}{2}$. The $4^{th}$ term $(T_4)$ corresponds to $r = 3$: $T_4 = \sqrt[3]{4} 	imes \frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)}{3!} (-\frac{3x}{2})^3$ $T_4 = \sqrt[3]{4} 	imes \frac{\frac{2}{3} 	imes (-\frac{1}{3}) 	imes (-\frac{4}{3})}{6} 	imes (-\frac{27x^3}{8})$ $T_4 = \sqrt[3]{4} 	imes \frac{8/27}{6} 	imes (-\frac{27x^3}{8})$ $T_4 = \sqrt[3]{4} 	imes \frac{8}{162} 	imes (-\frac{27x^3}{8}) = -\frac{\sqrt[3]{4}}{6} x^3$.

If $|x| < \frac{2}{3}$ then the $4^{th}$ term in the expansion of $(3x - 2)^{2/3}$ is

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