<s>[INST] 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.
 [/INST] For $\,(\sqrt[3]4-1)^2\,$, using $(a-b)(a^2+ab+b^2)=a^3-b^3$ gives $\,\displaystyle
\left(\sqrt[3]{4}-1\right)\left(\sqrt[3]{4^2}+\sqrt[3]{4}+1\right) = 3
$.
Therefore $\,\displaystyle
\sqrt[3]{4}-1=\frac{3}{2 \sqrt[3]{2}+\sqrt[3]{4}+1} = \frac{3}{\left(\sqrt[3]{2}+1\right)^2}\,$.
Then, using $\,a^2-b^2=(a-b)(a+b)\,$ gives the following, which is equivalent to the one posted:
$$\require{cancel}
\left(\sqrt[3]4-1\right)^2=\frac {3\left(\sqrt[3]4-1\right)}{\left(\sqrt[3]{2}+1\right)^2} = \frac{3\left(\sqrt[3]2-1\right)\bcancel{\left(\sqrt[3]2+1\right)}}{\left(\sqrt[3]{2}+1\right)^\bcancel{2}} = 3 \cdot \frac{\sqrt[3]2-1}{\sqrt[3]2+1}
$$
For $\,\left(\sqrt[3]{2}-1\right)^2\,$, using similar identities:
$$\sqrt[3]4-2 \sqrt[3]{2}+1=-\left(\sqrt[3]{4^2}-\sqrt[3]{4}+1\right)+2 = \dfrac{-5}{\sqrt[3]{4}+1}+2=\dfrac{2\sqrt[3]{4}-3}{\sqrt[3]{4}+1}\cdot\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\dfrac{4 - 3 \sqrt[3]{2}}{2+\sqrt[3]{2}}\,$$
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