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. 

Fun video!
Much time was spent on finding $abc=1/6$.
Alternative method for this:
$$\begin{align}a^2+b^2&=2-c^2 \Rightarrow \\
(a+b)^2-2ab&=2-c^2 \Rightarrow \\
(1-c)^2-2ab&=2-c^2 \Rightarrow \\
ab&=c^2-c-\frac12 \Rightarrow \\
abc&=c^3-c^2-\frac c2 \end{align}$$
Similarly:
$$abc=a^3-a^2-\frac a2\\
abc=b^3-b^2-\frac b2$$
Now adding them up:
$$3abc=(a^3+b^3+c^3)-(a^2+b^2+c^2)-\frac12(a+b+c)=3-2-\frac12 \Rightarrow abc=\frac16.$$
In fact, you can find other terms as well:
$$ab+bc+ca=(a^2+b^2+c^2)-(a+b+c)-\frac32=2-1-\frac32=-\frac12;\\
a^2b^2+b^2c^2+c^2a^2=ab(c^2-c-\frac12)+bc(a^2-a-\frac12)+ca(b^2-b-\frac12)=\\
abc(a+b+c)-3abc-\frac12(ab+bc+ca)=\\
\frac16-\frac12+\frac14=-\frac1{12}$$
Hence:
$$a^5+b^5+c^5=(a^2+b^2+c^2)(a^3+b^3+c^3)-(a^2b^2+b^2c^2+c^2a^2)+abc(ab+bc+ca)=\\
2\cdot 3-(-\frac1{12})+\frac16\cdot (-\frac12)=6.$$
