<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] Using Newton's identities
$$
\begin{aligned}
 e_{1}&=p_{1}\\
2e_{2}&=e_{1}p_{1}-p_{2}\\
3e_{3}&=e_{2}p_{1}-e_{1}p_{2}+p_{3}\\
4e_{4}&=e_{3}p_{1}-e_{2}p_{2}+e_{1}p_{3}-p_{4}\\
5e_{5}&=e_{4}p_{1}-e_{3}p_{2}+e_{2}p_{3}-e_{1}p_{4}+p_{5}\\
\end{aligned}
$$
with $p_1=1,p_2=2,p_3=3,e_4=0, e_5=0$, we get $p_5 = 6$.
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