= Solution
For $f(x)=x^{1/3}$, the required <derivatives> are
$$
f'(x)=\frac13x^{-2/3},\quad f''(x)=-\frac29x^{-5/3},\quad
f^{(3)}(x)=\frac{10}{27}x^{-8/3}.
$$
At $a=8$ these give $f(8)=2$, $f'(8)=1/12$, and $f''(8)=-1/144$. Setting $x-a=1$, the three-term <Taylor polynomial> yields
$$
\boxed{9^{1/3}\approx2+\frac1{12}-\frac1{288}=\frac{599}{288}\approx2.07986111.}
$$
By the <Taylor theorem with Lagrange remainder>, the signed error is
$$
R_3=\frac{f^{(3)}(\xi)}{6}=\frac5{81}\xi^{-8/3},\qquad 8<\xi<9.
$$
It is positive, and $\xi^{-8/3}<8^{-8/3}=1/256$. Consequently
$$
\boxed{0<9^{1/3}-\frac{599}{288}<\frac5{20736}<\frac1{4000}.}
$$
The last comparison is exact because $5\cdot4000=20000<20736$. Thus the estimate is an underestimate with the required rigorous <absolute error> bound.
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