Find value of $x^{5} + y^{5} + z^{5}$ given the values of $x + y + z$, $x^{2} + y^{2} + z^{2}$ and $x^{3} + y^{3} + z^{3}$ If$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$
Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations , substitute $z=1-x-y$ , but things got messy nothing seems to work out .

Another way.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$,where $v^2$ can be negative,  and $xyz=w^3$.
Thus, $u=\frac{1}{3}$ and since
$$2=x^2+y^2+z^2=9u^2-6v^2,$$
we obtain $$v^2=-\frac{1}{6}.$$
Also, since $$3=x^3+y^3+z^3=27u^3-27uv^2+3w^3=1+\frac{3}{2}+3w^3,$$ we get $$w^3=\frac{1}{6}.$$
Id est, $$x^5+y^5+z^5=243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3=6.$$
I used the following identities, which easy to prove and which we always use in the $uvw$'s method.
About $uvw$ see here https://artofproblemsolving.com/community/c6h278791
$$x^2+y^2+z^2=9u^2-6v^2,$$
$$x^3+y^3+z^3=27u^3-27uv^2+3w^3$$ and
$$x^5+y^5+z^5=243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3.$$
There are also:
$$\sum_{cyc}(x^2y+x^2z)=9uv^2-3w^3,$$
$$\sum_{cyc}x^2y^2=9v^4-6uw^3$$ and more
