Can Rational Triples Simplify Nested Cubic Roots?

[anemone]

Here is this week's POTW:

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Find a triple of rational numbers $(x,\,y,\,z)$ such that \(\displaystyle \sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\).

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[anemone]

Congratulations to Opalg for his correct solution, which you can find below::)

Spoiler

To start with, it seems very likely that the solution will be of the form $\sqrt[3]{\sqrt[3]2 - 1} = a + b\sqrt[3]2 + c\sqrt[3]4$, for some rational numbers $a,b,c.$ To save writing, I will write $\lambda$ for $\sqrt[3]2$. So we want to find rational numbers $a,b,c$ such that $\sqrt[3]{\lambda - 1} = a + b\lambda + c\lambda^2,$ where $\lambda^3 = 2.$ In that case, $$\lambda - 1 = (a + b\lambda + c\lambda^2)^3 = a^3 +2b^3 + 4c^3 + 12abc + 3\lambda(2ac^2 + a^2b + 2b^2c) + 3\lambda^2(a^2c + ab^2 + 2bc^2).$$ Comparing coefficients of powers of $\lambda$ on both sides, we want $$\begin{aligned}a^3 +2b^3 + 4c^3 + 12abc &= -1,\qquad(1)\\ 2ac^2 + a^2b + 2b^2c &= \tfrac13,\qquad(2) \\ a^2c + ab^2 + 2bc^2 &= 0.\qquad(3)\end{aligned}$$ Those equations do not look easy to solve. But (3) seemed like the one to start with, since there is a $0$ on the right side. It is clear from (3) that $a,b,c$ cannot all be positive, so I wondered whether there might be a solution with $a = 1$ and $b = -1.$ If so, then (3) becomes $2c^2 - c - 1 = 0,$ a quadratic with solutions $c = 1$ and $c = -\frac12$. The solution $c=1$ is not promising. But if we put $a=1$, $b=-1$ and $c = -\frac12$ in the left side of (1) and (2) then we get $$\begin{aligned}a^3 +2b^3 + 4c^3 + 12abc &= \tfrac92,\qquad(4)\\ 2ac^2 + a^2b + 2b^2c &= -\tfrac32,\qquad(5).\end{aligned}$$ The numbers on the right side of (4) and (5) are not the same as those in (1) and (2). But in fact we have hit the jackpot here, because in each case the right side of (4) and (5) is the same multiple (namely $-\frac92$) of the corresponding number in (1) and (2). It follows that $\lambda - 1 = -\frac29(a + b\lambda + c\lambda^2)^3.$ So when we take cube roots and substitute for the values of $a,b,c$ and $\lambda$, we get $$\sqrt[3]{\sqrt[3]2 - 1} = \sqrt[3]{-\frac29} + \sqrt[3]{\frac49} + \sqrt[3]\frac19.$$

Alternate solution:

Spoiler

Let \(\displaystyle a=\sqrt[3]{\sqrt[3]{2}-1}\) and \(\displaystyle b=\sqrt[3]{2}\).

This implies \(\displaystyle a=\sqrt[3]{b-1}\) and \(\displaystyle b^3=2\).

Note that

\(\displaystyle b^3=2\\b^3-1=1\\(b-1)(b^2+b+1)=1\)

and

\(\displaystyle \begin{align*}b^2+b+1&=\frac{3b^2+3b+3}{3}\\&=\frac{2+3b^2+3b+1}{3}\\&=\frac{b^3+3b^2+3b+1}{3}\\&=\frac{(b+1)^3}{3}\end{align*}\)

which implies

\(\displaystyle a^3=b-1=\frac{1}{b^2+b+1}=\frac{3}{(b+1)^3}\implies a=\frac{\sqrt[3]{3}}{b+1}---(1)\)

On the other hand,

\(\displaystyle 3=2+1=b^3+1=(b+1)(b^2-b+1)\)

from which it follows that

\(\displaystyle \frac{1}{b+1}=\frac{b^2-b+1}{3}---(2)\)

Combining equations (1) and (2), we get

\(\displaystyle a=\sqrt[3]{\frac{1}{9}}\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)\)

Consequently, \(\displaystyle (x,\,y,\,z)=\left(\frac{4}{9},\,-\frac{2}{9},\,\frac{1}{9}\right)\) is the desired triples.