The expression frac{1}{(x^2 + frac{1}{x})^{4/ | vedclass

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Question

(D) The given expression is $\frac{1}{(x^2 + \frac{1}{x})^{4/3}} = \frac{1}{(x^3(1 + \frac{1}{x^3}))^{4/3}} = \frac{1}{x^4(1 + \frac{1}{x^3})^{4/3}} =

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$|x| > 1$

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(D) The given expression is $\frac{1}{(x^2 + \frac{1}{x})^{4/3}} = \frac{1}{(x^3(1 + \frac{1}{x^3}))^{4/3}} = \frac{1}{x^4(1 + \frac{1}{x^3})^{4/3}} = x^{-4}(1 + \frac{1}{x^3})^{-4/3}$.For the binomial expansion $(1 + z)^n$ to be valid,the condition $|z| Here,$z = \frac{1}{x^3}$,so we require $|\frac{1}{x^3}| This implies $|x^3| > 1$,which simplifies to $|x| > 1$.

The expression $\frac{1}{(x^2 + \frac{1}{x})^{4/3}}$ can be expanded by the binomial theorem if:

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