Fast! Result : Let $ p$ in the form $ 6k + 5$ then $ \{1^3,..,(p - 1)^3\}$ is an reduce reside mod p. Now we chose $ x = \frac {a}{p},y = \frac {b}{p},z = \frac {c}{p}$ Chose $ a^3 + b^3\equiv c^3 (\mod p)$ and $ (c,p) = 1$ then $ (x,y,z)$ is solution of equation. Other can chose $ p = 5$ $ x = \frac {125k + 1}{5},y = \frac {125k + 2}{5},z = \frac {125k + 4}{5}$

TTsphn wrote: Fast! Result : Let $ p$ in the form $ 6k + 5$ then $ \{1^3,..,(p - 1)^3\}$ is an reduce reside mod p. Now we chose $ x = \frac {a}{p},y = \frac {b}{p},z = \frac {c}{p}$ Chose $ a^3 + b^3\equiv c^3 (\mod p)$ and $ (c,p) = 1$ then $ (x,y,z)$ is solution of equation. Other can chose $ p = 5$ $ x = \frac {125k + 1}{5},y = \frac {125k + 2}{5},z = \frac {125k + 4}{5}$ I think some thing is wrong in this solution...actually we must chose $ a,b,c$ such that $ a^3 + b^3\equiv c^3 (\mod p^3)$ my solution: Lemma.let $ p$ be a prime number in the form $ 3k + 2$ and $ a$ is integer and coprime with $ p$ then there exist integer $ t$ such that: $ a\equiv t^3 (\mod p^3)$ proof.note that $ \phi(p^3) = p^2(p - 1)$ therefore $ (\phi(p^3),3) = 1$ so there exist $ x,y\in \mathbb{N}$ such that $ 3y = \phi(p^3)x + 1$ take $ t = a^y$... $ t^3 = (a^y)^3 = a^{\phi(p^3)x + 1} = a\times (a^{\phi(p^3)})^x\equiv a (\mod p^3)$ therefore if we chose $ m,n,k$ such that be coprime with $ p$ and $ m + n = k$ then obviusly there exist $ a,b,c$ such that $ a^3 \equiv m (\mod p^3) , b^3 \equiv n (\mod p^3) , c^3 \equiv k (\mod p^3)$ and therefore $ x = \frac {a}{p},y = \frac {b}{p},z = \frac {c}{p}$ work... for example we find some solution for $ p = 5$: let $ m = 1,n = 8,k = 9$ therefore obviously $ a = 1,b = 2$ and for find $ c$ such that $ c^3 \equiv 9 (\mod 5^3)$note that: $ c^3 \equiv 9 (\mod 5) \Rightarrow c = 5k - 1 \\ c^3 \equiv 9 (\mod 5^2) \Rightarrow 125k^3 - 3\times25k^2 + 3\times5k - 1\equiv 9(\mod 5^2) \Rightarrow k = 5t - 1 \Rightarrow c = 25t - 6 \\ c^3 \equiv 9 (\mod 5^3) \Rightarrow 25^3t^3 - 3\times25^2t^2\times6 + 3\times25t\times36 - 216\equiv 9(\mod 5^2)\Rightarrow t = 5s + 3 \Rightarrow c = 125s + 69$ therefore $ x = \frac {1}{5},y = \frac {2}{5},z = \frac {69}{5}$ is a solution...and obviusly for every integer k , $ x = \frac {125k + 1}{5},y = \frac {125k + 2}{5},z = \frac {125k + 69}{5}$ is solution...

Perhaps one of the easiest solutions is to take $x = \frac{1}{4}, y = \frac{7}{4}$, and $z = \frac{3}{2} + 4n$ with $n \in \mathbb{N}$. We then get the following: \begin{align*} \left\{x^3\right\} &= \frac{1}{64}\\ \left\{y^3\right\} &= \left\{\frac{343}{64}\right \} = \left\{5 + \frac{23}{64}\right\} = \frac{23}{64} \\ \left\{z^3\right\} &= \left\{\frac{27}{8} + 27n + 72n^2 + 64n^3\right\} = \left\{\frac{27}{8}\right\} = \frac{3}{8} \end{align*} And now we indeed see $\{x^3\} + \{y^3\} = \frac{1}{64} + \frac{23}{64} = \frac{3}{8} = \{z^3\}$.