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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.

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  • $\begingroup$ You might also want to add the tag: nested radicals because when you take the square root of both sides, the right-hand side in your identity essentially turns into a nested radical.$$\sqrt{3\left(\frac {2-\sqrt[3]{4}}{2+\sqrt[3]{4}}\right)}=\sqrt[3]4-1$$ $\endgroup$ – Crescendo Jun 15 '18 at 20:28
  • $\begingroup$ @Crescendo Thanks Crescendo, will do. $\endgroup$ – Frank Jun 15 '18 at 20:38
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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 $(1)$.

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  • $\begingroup$ I feel like I'm being dumb, but how did you get that so quickly? $\endgroup$ – Frank Jun 15 '18 at 20:37
  • $\begingroup$ @Frank: simple computer search. $\endgroup$ – Oleg567 Jun 15 '18 at 20:39
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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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