Where's Number One?

Let $S_1= a+b+c$ , $S_2 = ab+bc+ca$ , $S_3 = abc$ and $P_n = a^n+b^n+c^n$ , where $n = 1,2,3$ .... Using Newton's Sums method, we have:

$\begin {cases} P1 = S_1 & & \\ P_2 = S_1^2 - 2S_2 & = 26 &...(1) \\ P_3 = 26S_1 - S_1S_2 + 3S_3 & = -129 & ...(2) \\ P_4 = -129S_1 - 26S_2 + S_1S_3 & = 650 &...(3) \end {cases}$

Eqn 2 $\times\ S_1$ - Eqn 3 $\times 3$ :

$\begin {cases} 26S_1^2 - S_1^2S_2 + 3S_1S_3 & = -129S_1 \\ -387S_1 - 78S_2 + 3S_1S_3 & = 1950 \end {cases}$

$\Rightarrow 26S_1^2 +(78-S_1^2)S_2 + 516S_1 = - 1950\quad ...(4)$

From eqn 1, we have $2S_2 = S_1^2-26$ . Substituting this in eqn 4 $\times 2$ :

$\Rightarrow 52S_1^2 +(78-S_1^2)(S_1^2-26) + 1032S_1 = - 3900$

$\quad 52S_1^2 +(-S_1^4+104S_1^2-2028)+ 1032S_1 = - 3900$

$\quad -S_1^4 + (52+104)S_1^2 + 1032S_1 -2028 + 3900 = 0$

$\quad S_1^4 - 156S_1^2 - 1032S_1 -1872 = 0$

$\quad (S_1+6)(S_1^3 - 6S_1^2 - 120S_1 -312) = 0$

$\Rightarrow S_1 = -6\quad \Rightarrow S_2 = 5\quad \Rightarrow S_3 = -1$

Now we continue to find $P_5$ and $P_6$ .

$P_5 = a^5+b^5+c^5$

$\quad \space \space = 650(S_1) + 129(S_2) + 26(S_3)$

$\quad \space \space = 650(-6) + 129(5) + 26(-1) = -3900+645 -26 = -3281$

$P_6 = a^6+b^6+c^6$

$\quad \space \space = -3281(-6) - 650(5) + 129(-1) = 19686-3250+129 = 16565$

Therefore, the required answer $a^6+b^6+c^6 \pmod {1000} \equiv 16565 \pmod {1000} \equiv \boxed {565} \pmod {1000}$

Why performing and therefore repeating well known algorithms in time expanding manner when it is possible to apply mathsoftware for example XMaxima? " solve([a^2+b^2+c^2=26,a^3+b^3+c^3=-129,a^4+b^4+c^4=650],[a,b,c]); " and final calculation of the expression wished will then deliver the solution faster with much higher efficiency!

And so you even have time for visiting the christmas market or doing similar things. ;-)