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

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

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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 $6^{-1/3} (1 - \frac{3x}{6})^{-1/3} = 6^{-1/3} (1 - \frac{x}{2})^{-1/3}$.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]$.

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

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