Can i factor this expression: $x^3+y^3+z^3$ I have the following numerical expression, which is exactly equal to $1$

Text version:
(-(7 2^(2/3))/38 + 3^(2/3)/19 + (23 5^(2/3))/95 + (5 6^(1/3))/19 + 3/38 (5^(1/3) 6^(2/3)) - 10^(1/3)/19 - 3/190 (3^(1/3) 10^(2/3)) - (4 15^(1/3))/19 - 6/95 (2^(1/3) 15^(2/3)))*(2^1/3+3^1/3+5^1/3)

Image version:

according to this fact, $x^3+y^3+z^3$ would  be factorized at all and first factor of it should be $x+y+z$ if we set
$
x^3=2,y^3=3,z^3=5
$


If so then how to factor $x^3+y^3+z^3$ ?


PS. I tried to divide $x^3+y^3+z^3$ by $x+y+z$ and failed (got remainder).

Your expression can be verified by the following identity:
\begin{align}
  & \; \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c} \\
  = & \:
  \dfrac
   {(a+b+c)^3-27abc}
   {\left(
       \sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}-
       \sqrt[3]{bc}-\sqrt[3]{ca}-\sqrt[3]{ab} \,
    \right)
    \left[
       (a+b+c)^2+
       3(a+b+c)\sqrt[3]{abc}+
       9\sqrt[3]{a^2 b^2 c^2}
    \right]}
\end{align}
Note that $$(a,b,c)=(2,3,5) \implies (a+b+c)^3-27abc=190$$
Your first factor is just rationalization of the denominator for
$$\frac{1}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}}$$
