Find rational numbers $a,b,c$ satisfying $(2^{1/3}-1)^{1/3} = a^{1/3}+b^{1/3}+c^{1/3}$. 
$\textbf{Problem} $ Find rational numbers $a,b,c$ satisfying
  \begin{align*}
(2^\frac13-1)^\frac13 = a^\frac13+b^\frac13+c^\frac13
\end{align*}

My Attempt: I try to $2^\frac13-1 = (a^\frac13+b^\frac13+c^\frac13)^3$ and compare with rational numbers and irrational numbers in LHS and RHS. 
Any help is appreciated...
Thank you!
Update: I found the answer. I want to find rational numbers $a,b,c$ without assumption $a=1/9,b=-2/9,c=4/9$. Thus,  I found the identity:
$$\sqrt[3]{m^3-n^3+6m^2n+3mn^2-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}=\\ \sqrt[3]{m^2(m+n)}-\sqrt[3]{mn^2}-\sqrt[3]{(m+n)^2n}$$
1) How to get the identity?
2) I want to know about uniqueness $(a,b,c)$ satisfying $(2^\frac13-1)^\frac13=a^\frac13+b^\frac13+c^\frac13$
 A: Set $t:=\sqrt[3]{2}$.  Then, $t^3=2$ and $t^3-1=1$, whence
$$t-1=\frac{t^3-1}{t^2+t+1}=\frac{1}{t^2+t+1}=\frac{3}{3t^2+3t+3}=\frac{3}{t^3+3t^2+3t+1}=\frac{3}{(t+1)^3}\,.$$
Therefore,
$$\sqrt[3]{t-1}=\frac{\sqrt[3]{3}}{t+1}\,.$$
Now, $t^3+1=3$, so
$$\sqrt[3]{t-1}=\frac{\sqrt[3]{3}\left(t^2-t+1\right)}{t^3+1}=\frac{\sqrt[3]{3}\left(t^2-t+1\right)}{3}=\frac{t^2-t+1}{\sqrt[3]{9}}\,.$$
Plugging in $t=\sqrt[3]{2}$, we obtain
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{4}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac19}\,.$$
In fact, this is the only possible way to write $\sqrt[3]{\sqrt[3]{2}-1}=\frac{1}{3}\sqrt[3]{12}-\frac{1}{3}\sqrt[3]{6}+\frac{1}{3}\sqrt[3]{3}$ as a sum $$\sum_{j=1}^n\,x_j\,\sqrt[r_j]{d_j}\,,$$ where $x_1,x_2,\ldots,x_n\in\mathbb{Q}\setminus\{0\}$, $r_1,r_2,\ldots,r_n\in\mathbb{Z}_{>0}$, and for $j=1,2,\ldots,n$, each $d_j\neq0$ is an $r_j$-power-free integer (with $d_j=1$ iff $r_j=1$, and with $d_j>1$ if $r_j>1$ is odd).  This is because radicals are linearly independent over $\mathbb{Q}$.  See a proof here.  (As a consequence, $a$, $b$, and $c$ are unique, up to permutation.)
