Determining rational functions that simplify to a simple expression 
Question: Is there a way to find, if possible, another way to write $(\sqrt[3]2-1)^2$ and $(\sqrt[3]4-1)^2$, in the form of $\frac {a+b}{c+d}$?

What I meant was that, let's take the second expression $(\sqrt[3]4-1)^2$ as an example. The expression is also equal to$$(\sqrt[3]4-1)^2=3\left(\frac {2-\sqrt[3]4}{2+\sqrt[3]4}\right)$$
You can check this with Wolfram Alpha. However, I'm not sure how to do that with $(\sqrt[3]2-1)^2$ and what the work is. I feel like this should be such a basic thing to do, but I'm struggling with where to begin.

As for $(\sqrt[3]{2}-1)^2$:
$$
\dfrac{4-3\sqrt[3]{2}}{2+\sqrt[3]{2}} = (\sqrt[3]{2}-1)^2. \tag{1}
$$
We can easily check it: denote $x = \sqrt[3]{2}$, then $$(x-1)^2(2+x)=(x^2-2x+1)(2+x)= \\ 2x^2-4x+2+x^3-2x^2+x=\\ x^3-3x+2 = \\ 4-3x.$$

Let's try to find $(x-1)^2$ in the form 
$$
\dfrac{a+bx}{c+dx}, \qquad a,b,c,d\in\mathbb{Z}\tag{2}$$ manually; and prove this way that $(1)$ is unique representation for $(x-1)^2$ (where $x=\sqrt[3]{2}$) in the form $(2)$ (of course, if avoid scaling of $a,b,c,d$).
Then 
$$
(x-1)^2(c+dx)=a+bx; \\
(x^2-2x+1)(c+dx)=a+bx; \\
dx^3+(c-2d)x^2+(d-2c)x+c=bx+a; \\
(c-2d)x^2+(d-2c)x+(c+2d)=bx+a.
$$
Since $1,\sqrt[3]{2}, \sqrt[3]{4}$ are linearly independent over $\mathbb{Q}$, we have system of equations:
$$
c-2d=0, \\
b=d-2c, \\
a=c+2d.
$$
So, $c=2d$, $b=-3d$, $a=4d$. Setting $d=1$, we'll get $(1)$.
