Find the rational number of a, b, c, solving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+ \sqrt[3]{b}+\sqrt[3]{c}$

Question

I try as following let \begin{eqnarray} x= \sqrt[3]{a} \\ y= \sqrt[3]{b} \\ z= \sqrt[3]{c} \\ x+y+z = \sqrt[3]{\sqrt[3]{2}-1 }\\ \end{eqnarray} We know that \begin{equation} x^3+y^3+z^3 = (x+y+z)^3-3(x+y)(x+z)(y+z) \end{equation} that turns out to be \begin{equation} (x+y+z-1)(x+y+z)(x+y+z-1)=3(x+y)(x+z)(y+z) \end{equation} plug in $x+y+z$, we will get \begin{equation} \sqrt[3]{2}-1 - \sqrt[3]{\sqrt[3]{2}-1 } = 3(x+y)(x+z)(y+z) \end{equation} From now I stuck to solve for the rational numbers of $a,b,c$. Could anyone show me the way how to continue to solve it?

Answer 1

An observation:

\begin{align*} (\sqrt[3]{4}-\sqrt[3]{2}+1)^3&=\frac{(\sqrt[3]{8}+1)^3}{(\sqrt[3]{2}+1)^3}\\ &=\frac{27}{2+3\sqrt[3]{4}+3\sqrt[3]{2}+1}\\ &=\frac{9}{\sqrt[3]{4}+\sqrt[3]{2}+1}\\ &=\frac{9(\sqrt[3]{2}-1)}{\sqrt[3]{8}-1}\\ &=9(\sqrt[3]{2}-1)\\ \left(\sqrt[3]{\frac{4}{9}}+\sqrt[3]{\frac{-2}{9}}+\sqrt[3]{\frac{1}{9}}\right)^3&=\sqrt[3]{2}-1 \end{align*}

Answer 2

I try this: $y=\sqrt[3]{2}$,$y^3=2$, $x= \sqrt[3]{\sqrt[3]{2}-1}$ Therefore, $x^3=y-1$ $y^3-1=(y-1)(y^2+y+1)=1$ Because $y^2+y+1=\frac{3y^2+3y+3}{3}=\frac{y^3+3y^2+3y+1}{3}=\frac{(y+1)^3}{3}$ and $y^3+1=3=(y+1)(y^2-y+1)$

Hence

$x^3=y-1=\frac{1}{y^2+y+1}=\frac{3}{(y+1)^3}=\frac{1}{9}(y^2-y+1)$

Therefore, based on the definition of $y$, we can obtain $a=\frac{4}{9}$,$b=-\frac{2}{9}$, and $c=\frac{1}{9}$

Answer 3

Here is an alternative to denesting $(2^{1/3}-1)^{1/3}$. First, set $x^3=2$ so that$$x^3-1=1$$Factoring the left-hand side and isolating $x-1$ gives$$x-1=\frac 1{1+x+x^2}=\frac 3{3+x+3x^2}=\frac 3{(1+x)^3}$$Multiply both sides by $9$ to complete the cube$$9(x-1)=\left(\frac 3{1+x}\right)^3$$Cube root both sides and set $x=\sqrt[3]{2}$ gives$$\sqrt[3]{\sqrt[3]2-1}=\frac {3}{1+\sqrt[3]2}=1-\sqrt[3]2+\sqrt[3]4$$Hence$$\sqrt[3]{\sqrt[3]2-1}\color{blue}{=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}}$$A similar technique can be done to show that$$\sqrt[3]{7\sqrt[3]{20}-1}=\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}$$

Answer 4

A systematic approach is to apply the denesting formula

$$\sqrt[3]{\sqrt[3]{A}-B} = \sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 } $$ where $x_1$, $x_2$ and $x_3$ are the roots of the cubic equation $$x^3 + \frac{B+2C}3x^2 - \frac{(B-C)(2B+C)}{27}x+ \frac{(B-C)^3}{729}=0$$ with $C =\sqrt[3]{B^3-A}$. Thus, to denest $\sqrt[3]{\sqrt[3]{2}-1}$, solve $$x^3 -\frac13 x^2-\frac2{27}x + \frac8{729}=0\implies (x_1, x_2, x_3)= (\frac19,-\frac29,\frac49)$$

to obtain

\begin{align} \sqrt[3]{\sqrt[3]{2}-1}&=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49} \end{align}

The general formula given above can be used as well to denest other nested cubic radicals, for instance

$$\sqrt[3]{21\sqrt[3]{6}-17}=\sqrt[3]{18}+\sqrt[3]{4}-\sqrt[3]{3}$$