Can i factor this expression: $x^3+y^3+z^3$

Question

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).

Answer 1

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

$$\frac{1}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}}$$

Answer 2

You have the identity:

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$$

As far as I know, this is the best you can do, though I'd be happy if someone proves me wrong.

Answer 3

By Eisenstein's criterion and Gauss' lemma, $x^3+y^3+z^3$ is irreducible in $k[x,y,z]$ for any field $k$ not of characteristic $3$. So, for example, it is irreducible in $\mathbb C[x,y,z]$.