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Question: Answered & Verified by Expert
$\frac{1}{\sqrt[3]{6-3 x}}=$
MathematicsBinomial TheoremJEE Main
Options:
  • A $6^{1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots.\right]$
  • B $6^{-1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots.\right]$
  • C $6^{1 / 3}\left[1-\frac{x}{6}+\frac{2 x^2}{6^2}-\ldots.\right]$
  • D $6^{-1 / 3}\left[1-\frac{x}{6}+\frac{2 x^2}{6^2}-\ldots\right]$
Solution:
2625 Upvotes Verified Answer
The correct answer is: $6^{-1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots.\right]$
$\begin{aligned} & \frac{1}{(6-3 x)^{1 / 3}}=(6-3 x)^{-1 / 3}=6^{-1 / 3}\left[1-\frac{x}{2}\right]^{-1 / 3} \\ & =6^{-1 / 3}\left[1+\left(-\frac{1}{3}\right)\left(-\frac{x}{2}\right) x+\frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2.1}\left(-\frac{x}{2}\right)^2+\ldots .\right] \\ & =6^{-1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots .\right]\end{aligned}$

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