<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] Let's start with the basic symmetric expressions: $ab+bc+ca$ and $abc$. You can refer to giannispapav's answer for details, which shows that
$$ab+bc+ca = -1/2, abc = 1/6.$$
With that, Vieta's formulas implies that $a,b,c$ satisfy:
$$ x^3 -x^2 - x/2  -1/6=0,\tag{1}$$
Or
$$x^3 = x^2 + x/2 + 1/6.$$
That means, for $x$ equals $a,b,c$,
$$x^4 = x^3 + x^2/2 + x/6,$$
and
$$x^5 = x^4 + x^3/2 + x^2/6.$$
Adding the two equations above, we have
$$x^5 = \frac32x^3 + \frac23x^2 + \frac16x.$$
Now replace $x$ as $a,b,c$ and add them all up, we have
$$a^5+b^5+c^5 = \frac32(a^3+b^3+c^3) + \frac23(a^2+b^2+c^2) + \frac16(a+b+c).$$

Note: if you feels that
$$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$$
is too complicated to verify, then Vieta's formulas is the way to go. That is, replace $a,b, c$ in Equation $1$ and add them up, where $1/6$ is indeed $abc$ as in Vieta's formula.
</s>