Bashing Polynomial

The problem can be solved by Newton's Sums method.

Let $P_n = a^n+b^n+c^n$ , where $n=1,2,3...$ and:

$S_1=a+b+c = 2 \quad \quad S_2 = ab+bc+ca = 3\quad \quad S_3 = abc = -1$ .

Then, $P_n$ are given as follows:

$\begin {cases} P_1 = S_1 & &= 2 \\ P_2 = S_1P_1 - 2S_2 & = 2(2)-2(3) & = -2 \\ P_3 = S_1P_2 - S_2P_1 +3S_3 & = 2(-2)-3(2) + 3(-1) & = -13 \\ P_4 = S_1P_3 - S_2P_2 +S_3P_1 & = 2(-13)-3(-2) -1(2) & = -22 \\ P_5 = S_1P_4 - S_2P_3 +S_3P_2 & = 2(-22)-3(-13) -1(-2) & = -3 \end {cases}$

Therefore,

$a^5+b^5+c^5 - (a^4+b^4+c^4) = P_5 - P_4 = -3 - (-22) = \boxed {19}$

$Some~ thing~ from ~fundamentals.\\ By ~Vieta's:-\\ ~~~~~~~~a+b+c=2.\\ ~~~~~ab+bc+ca=3.\\ ~~~~~~~~~~abc= - 1.\\ a^2+b^2+c^2= -2..........................a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)= ~-2.\\ a^2b^2+b^2c^2+c^2a^2=13................. a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=13.\\ a^4+b^4+c^4= - 22.......................a^4+b^4+c^4=(a^2+b^2+c^2)^2 - 2(a^2b^2+b^2c^2+c^2a^2)= - 22\\ a^5+b^5+c^5= - 3.........................a^5+b^5+c^5=(a+b+c)(a^4+b^4+c^4) - \{a(b^4+c^4c)+b(c^4+a^4)+c(a^4+b^4)\}\\ ...............................................=(a+b+c)(a^4+b^4+c^4) - \{ab(a^3+b^3)+bc(b^3+c^3)+ca(c^3+a^3)\}\\ ...............................................=(a+b+c)(a^4+b^4+c^4) - \{ab(-13 - c^3)+bc(-13 -a^3)+ca(-13 - b^3)\}\\ ...............................................=(a+b+c)(a^4+b^4+c^4) - \{-13( ab+bc+ca)-abc(c^2+a^2+b^2)\}\\ ..........................................................(2)(-22)~~~~~~~~~~~~ - \{-13(3)~~~~~~~~~~~~ -~~~~~~~(-1)(-2)\}~~~~~~= - 3\\ \therefore~~a^5+b^5+c^5-(a^4+b^4+c^4)= - 3 -( - 22)= \Large \color{#D61F06}{19}.$

we use newton sums. http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Sums