Factoring a quartic expression 
We have a function $$f(x)=x^{\frac{2}{3}}-\frac{2}{x^{\frac{1}{3}}}+1$$ and let line segment $AB$ be represented by $g(x)$ such that $$g(x)=x-4$$
  Find the greatest positive real solution representing the intersection between the two lines.

The first step is obviously to let them equal each other, and the other steps will follow logically, they are demonstrated here
$$x-4=x^{\frac{2}{3}}-\frac{2}{x^{\frac{1}{3}}}+1$$ $$x-5=\frac{x-2}{x^{\frac{1}{3}}}$$ $$x^{\frac{1}{3}}=\frac{x-2}{x-5}$$ $$x=\frac{x^3-6x^2+12x-8}{x^3-15x^2+75x-125}$$ $$x^4-15x^3+75x^2-125x=x^3-6x^2+12x-8$$ $$x^4-16x^3+81x^2-137x+8=0$$
and here I'm very much stuck. By the factor theorem we know that for $(x-a)$ to a factor, $a$ must be a factor of $8$, i.e. $1,2,4,-1,-2$ or $-4$, but I've tried every one of these values and haven't found a value of $a$ such that $(x-a)$ is a factor of the latest expression.
The graph of the two equations is given here, and the point of intersection clearly falls at the intersection of two grid lines, so I know the solution is an integer, but under examination conditions I obviously won't have a graphing tool this sophisticated, so I'm eager to know how to proceed from here algebraically.



Any help is appreciated, thank you.

You are very close, just haven't tried $a=8$ yet.
\begin{align}
f(x)&=x^4 -16x^3+81x^2-137x+8 \\
f(8) &= 8^4 - 16 \cdot8^3 + 81 \cdot8^2 - 137\cdot8+8 \\
&= 4096-8192+5184-1096+8\\
&= 0
\end{align}
Then we know $(x-8)$ is a factor of $f(x)$.
\begin{align}
f(x)&=(x-8)(x^3-8x^2+17x-1) \\
\end{align}
Used polynomial division to find $(x^3-8x^2+17x-1)$.
Let $p(x) = x^3-8x^2+17x-1$
You can then prove that $p(x)$ has no integer solutions. Your end answer should be $x=8$.
