<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] This answer is almost in the same spirit as @Quang Hoang's, but I hope this answer will add something. Let $$
P(z) = (z-a)(z-b)(z-c)=z^3-\sigma_1 z^2+\sigma_2 z-\sigma_3
$$ where $\sigma_1=a+b+c$, $\sigma_2=ab+bc+ca$ and $\sigma_3=abc$ by Vieta's formula. Note that for $z\in \{a,b,c\}$,
$$
z^{n+3} =\sigma_1 z^{n+2}-\sigma_2 z^{n+1}+\sigma_3 z^n,
$$ hence by summing over $z\in \{a,b,c\}$, we get recurrence relation
$$
s_{n+3}= \sigma_1 s_{n+2}-\sigma_2 s_{n+1}+\sigma_3 s_n 
$$ for $s_n = a^n+b^n+c^n$. Given the data, it can be easily noted that $$\sigma_1=1 ,\quad \sigma_2 =\frac 12 \left((a+b+c)^2-(a^2+b^2+c^2)\right)=-\frac 12.$$ And by plugging $n=0$, we obtain
$$
3=1\cdot 2+\frac 12\cdot 1 +\sigma_3 s_0=2.5 + \sigma_3s_0,
$$ so $\sigma_3=abc\ne 0$ and $s_0=a^0+b^0+c^0=3$. This gives $\sigma_3=\frac 1 6$, implying that
$$
s_{n+3}=s_{n+2}+\frac 12 s_{n+1}+\frac 1 6 s_{n},\quad \forall n\ge 0.
$$ Now $s_4 =\frac {25}{6}$ and $s_5=6$ follows from the initial data $(s_3,s_2,s_1)=(3,2,1)$.
Note : The theory of homogeneous linear difference equations is behind it.
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