= Solution
Use the standard <Fourier transform> on the whole real axis. The printed lower limit is missing a minus sign. For this integrable even function, its only obstruction to smoothness is the origin, so its high-frequency algebraic terms come from its local cusp:
$$
\frac1{1+|x|^3}=1-|x|^3+|x|^6-|x|^9+|x|^{12}-\cdots.
$$
Localize near zero with a smooth cutoff. The smooth polynomial terms, including $|x|^6=x^6$, contribute no algebraic cusp term; their localized transforms decay faster than any prescribed inverse power. The supplied half-line formula, interpreted away from $k=0$ in an Abel-regularized or <tempered distribution> sense, gives the <Fourier transform of an algebraic cusp>
$$
\mathcal F(|x|^p)(k)=\frac{2\Gamma(p+1)\cos[\pi(p+1)/2]}{|k|^{p+1}}\quad(k\ne0).
$$
For $p=3$ the cosine is $1$, and for $p=9$ it is $-1$. Both cusp coefficients in the local expansion are $-1$. Hence
$$
\boxed{\widehat f(k)=-\frac{12}{|k|^4}+\frac{2\,9!}{|k|^{10}}+O(|k|^{-16})}.
$$
The second coefficient is $2\,9!=725760$. The absence of a $|k|^{-7}$ term follows from the smooth even power $x^6$, not from neglecting an available correction.
One can verify the signs by <integration by parts> on $2\operatorname{Re}\int_0^\infty e^{-ikx}(1+x^3)^{-1}dx$. Its endpoint derivatives first have nonzero relevant values $f^{(3)}(0)=-3!$ and $f^{(9)}(0)=-9!$, which contribute twice those values divided by $(ik)^4$ and $(ik)^{10}$. Higher derivatives are integrable on the half-line, justifying the displayed remainder after further integrations. Equivalently these coefficients are the <Fourier decay from a derivative jump> at the origin.
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