Condition for using binomial theorem for negative index

Question

Prove that $(x^2 + \frac{1}{x})^{-4/3}$ can be expanded by binomial theorem if $|x|>1$.

Now I know that $(1+x)^n$ can be expanded for negative index if $|x|<1$.

How should I prove it here?

Answer 1

HINT Assuming $x \ne 0$,

$$ \left(x^2+\frac{1}{x}\right)^{-4/3} = x^{-8/3} \left(1 + \frac{1}{x^3}\right)^{4/3} $$

Answer 2

$$(x^2 + \frac{1}{x})^{-4/3} = [x^2(1+\frac{1}{x^3})]^{-4/3} = x^{-8/3}(1+x^{-3})^{-4/3} $$

$x^{-8/3}$ is just a factor, so we can ignore it. Imposing the conditions you quote ($n < 0$ and $|x| < 1$), the expansion is valid if $-\frac{4}{3} < 0$ and $|x^{-3}| < 1$. Now clearly the former is true, but we need to work out what values of $x$ satisfies the latter.

$$ |\frac{1}{x^3}| < 1 \implies 1 < |x^3|$$

If $|x| \le 1$ then $|x^3| \le 1$, which does not satisfy our criterion.

If $|x| > 1$ then $|x^3| > 1$, so this is the range for which the expansion is valid.