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Question

(A) The expression is $\left(125 x^2 - \frac{27}{x}\right)^{-2/3} = \left(\frac{125 x^3 - 27}{x}\right)^{-2/3} = \frac{x^{2/3}}{(125 x^3 - 27)^{2/3}}$

Vedclass · Accepted Answer

$\left(-\frac{3}{5}, \frac{3}{5}\right)$

Vedclass · Answer

(A) The expression is $\left(125 x^2 - \frac{27}{x}ight)^{-2/3} = \left(\frac{125 x^3 - 27}{x}ight)^{-2/3} = \frac{x^{2/3}}{(125 x^3 - 27)^{2/3}}$.For the binomial expansion $(1+z)^n$ to be valid,we require $|z| Rewriting the expression: $\left(-\frac{27}{x}ight)^{-2/3} \left(1 - \frac{125 x^3}{27}ight)^{-2/3}$.This expansion is valid when $|\frac{125 x^3}{27}| $|x^3| Thus,$x \in \left(-\frac{3}{5}, \frac{3}{5}ight)$ and $x 
eq 0$.

The set of all real values of $x$ for which the expansion of $\left(125 x^2-\frac{27}{x}\right)^{-2/3}$ is valid,is

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