Sum Of Roots

It can be solved using Newton's Sum method, which I think is easier.

If the four roots of the equation are $a,b,c,d$ . Then, by Vieta's formulas, we have

$\quad S_1 = a +b+c+d = 0 \quad \quad S_2 = ab+ac+ad+bc+bd+cd = -7$

$\quad S_3 = abc+abd+acd+bcd =-4 \quad \quad S_4 = abcd = -3$

Then $P_n = a^n+b^n+c^n+d^n$ , where $n=1,2,3$ ... is given by:

$P_1 = S_1 = 0$

$P_2 = S_1P_1 - 2S_2 = 0 -2(-7) = 14$

$P_3 = S_1P_2 - S_2P_1 + 3S_3 = 0 +7(0) +3(-4) = -12$

$P_4 = S_1P_3 - S_2P_2 + S_3P_1 -4S_4 = 0 +7(14)-3(0) + 4(-3) = 86$

$P_5 = S_1P_4 - S_2P_3 + S_3P_2 -S_4P_1 = 0 +7(-12)-3(14) + 4(0) = -140$

Therefore, $P_5 = a^5+b^5+c^5+d^5 = \boxed {-140}$