Find the binomial expansion of frac{1}{sqrt[3 | vedclass

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(B) We have $\frac{1}{(6 - 3x)^{1/3}} = (6 - 3x)^{-1/3}$.Taking $6$ as a common factor,we get $6^{-1/3} (1 - \frac{3x}{6})^{-1/3} = 6^{-1/3} (1 - \fra

Vedclass · Accepted Answer

$6^{-1/3} \left[ 1 + \frac{x}{6} + \frac{2x^2}{6^2} + \dots \right]$

Vedclass · Answer

(B) We have $\frac{1}{(6 - 3x)^{1/3}} = (6 - 3x)^{-1/3}$.Taking $6$ as a common factor,we get $6^{-1/3} (1 - \frac{3x}{6})^{-1/3} = 6^{-1/3} (1 - \frac{x}{2})^{-1/3}$.Using the binomial expansion $(1 - y)^{-n} = 1 + ny + \frac{n(n+1)}{2!} y^2 + \dots$ for $|y| $6^{-1/3} \left[ 1 + (\frac{1}{3})(\frac{x}{2}) + \frac{(\frac{1}{3})(\frac{4}{3})}{2} (\frac{x}{2})^2 + \dots \right]$$= 6^{-1/3} \left[ 1 + \frac{x}{6} + \frac{4/9}{2} \cdot \frac{x^2}{4} + \dots \right]$$= 6^{-1/3} \left[ 1 + \frac{x}{6} + \frac{2x^2}{36} + \dots \right]$$= 6^{-1/3} \left[ 1 + \frac{x}{6} + \frac{2x^2}{6^2} + \dots \right]$.

List-$I$	List-$II$
$(A)$ $(1-x)^{-n}$	$(i)$ $\frac{x}{x+1}$
$(B)$ $(1+x)^{-n}$	$(ii)$ $1-nx+\frac{n(n+1)}{2!}x^2-\dots$ if $\|x\| < 1$
$(C)$ If $x>1$,then $1+\frac{1}{x}+\frac{1}{x^2}+\dots$ is	$(iii)$ $1+nx+\frac{n(n+1)}{2!}x^2+\dots$ if $\|x\| < 1$
$(D)$ If $\|x\|>1$,then $1-\frac{2}{x^2}+\frac{3}{x^4}-\frac{4}{x^6}+\dots$ is	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

Find the binomial expansion of $\frac{1}{\sqrt[3]{6 - 3x}}$.

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