<s>[INST] Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$ 
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
 \text{then}\ a^5+b^5+c^5= \ ?$$

A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the video, the best I can do with this is relying on expansion formulas and substitution. As trivial a problem this is, the numerous trinomials and binomials with mixed terms makes it very, very tedious.
What is the quickest/shortest approach to this problem (meaning it doesn't need to be solved algebraically)? You don't have to type the entire solution out, I think if I'm given a good hint then I can take it from there. 
 [/INST] I think to use an identity must be the shortest way:-
${ (a^5+b^5+c^5)=(a^4+b^4+c^4)(a+b+c)-(a^3+b^3+c^3)(ab+bc+ca)+abc(a^2+b^2+c^2)}  $
If we square the 1st equation we get
$ {ab+bc+ca=\frac {-1}2}  $
On squaring the above equation we get 
$ {(ab)^2+(bc)^2+(ca)^2+abc(a+b+c)=\frac 14}  $
We also know
${ (a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab+bc+ca) }$
from here we get :-
$abc=\frac 16$
Substituting the known values 
We get :-
${ (ab)^2+(bc)^2+(ca)^2=\frac {-1}{12} } $
Now squaring the 2nd equation:-
${ (a^4+b^4+c^4)+2\big ((ab)^2+(bc)^2+(ca)^2)\big )=4 }$
From here we get 
$(a^4+b^4+c^4)=\frac{25}6$
Now we have all the values to be substituted in the identity that i mentioned therefore you get 
$ { (a^5+b^5+c^5)= (\frac{25}{6}\cdot 1) - (3 \cdot \frac {-1}2)+ (\frac 16 \cdot 2) }$ 
$\Rightarrow$$(a^5+b^5+c^5)=6$
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