Is it as hard as it looks?

I used the Newton's Sums to solve this.

Let: $\quad P_1 = a + b + c \quad P_2 = a^2+b^2+c^2 \quad P_3 = a^3+b^3+c^3 \quad ...$

$\quad \quad \quad S_1 = a+b+c \quad S_2 = ab+bc+ca \quad S_3 = abc$

Then,

$P_1 = S_1 = 6$

$P_2 = S_1 P_1 - 2S_2 = 78 = 6 \times 6 - 2S_2 \quad \Rightarrow S_2 = -21$

$P_3 = S_1 P_2 - S_2 P_1 + 3S_3 = 420 = 6 \times 78 + 21 \times 6 + 3S_3 \quad \Rightarrow S_3 = - 58$

$P_4 = S_1 P_3 - S_2 P_2 + S_3P_1 = 6\times 420 + 21\times 78 - 58\times 6 = 3810$

$P_5 = S_1 P_4 - S_2 P_3 + S_3P_2 = 6\times 3810 + 21\times 420 - 58\times 78 = 27156 = x$

Therefore, $\quad x \mod {50} = 27156 \mod {50} = \boxed{6}$

There is a famous recurrence formula. a^(n+3)+b^(n+3)+c^(n+3)=(a+b+c)(a^(n+2)+b^(n+2)+c^(n+2))-(ab+bc+ca)(a^(n+1)=b^(n+1)+c^(n+1))+ abc(a^(n)+b^(n)+c^(n)) The rest is just plain arithmetic.