I've got the answer by converting the limit to $(x^3-8x^2)^{1/3}-x+1$ and then applying a "rationalizing" factor using the identity $a^3-b^3 = (a-b)(a^2+b^2+ab)$ to get it in fractional form. From here, I divide the top and bottom of the fraction by $x^2$ to get the answer. However, this was algebraically tedious and I was wondering if anyone could think of a better, more elegant way.
On
Note that, as $\displaystyle x \to \infty$, \begin{align} x^{2/3}\left(x - 8\right)^{1/3} & = x\left(1 - {8 \over x}\right)^{1/3} \sim x\left[1 + {1 \over 3}\left(-\,{8 \over x}\right) + {1/3\left(1/3-1\right) \over 2}\left(-\,{8 \over x}\right)^{2}\right] \\[5mm] & = x - {8 \over 3} - {64 \over 9x} \end{align} such that \begin{align} x^{2/3}\left(x - 8\right)^{1/3} - x + 1 & \sim \color{red}{\large-\,{5 \over 3}} - {64/9 \over x}\qquad\mbox{as}\quad x \to \infty \end{align}
You can pull the finite constant $+1$ out. Then by the conjugate trinomial trick, the numerator becomes $x^2(x-8)-x^3=-8x^2$, and there will be three terms (not cancelling) of order $x^2$ in the denominator plus lower powers.
Hence $-\dfrac83+1$.
Tedious ?