Mathematical Enigma - IV

Let the three numbers be $a$ , $b$ and $c$ . Using Newton's sums or Newton's identities , we have:

$\begin{aligned} a + b + c & = 6 \\ a^2+b^2+c^2 & = (a+b+c)^2 - 2(ab+bc+ca) \\ 6^2 - 2(ab+bc+ca) & = 8 \\ \Rightarrow ab+bc+ca & = \frac{36-8}{2} = 14 \\ a^3+b^3+c^3 & = (a+b+c) (a^2+b^2+c^2-ab-bc -ca) +3abc \\ 6(8-14) + 3abc & = 5 \\ \Rightarrow abc & = \frac{36+5}{3} = \frac{41}{3} \\ a^4+b^4+ c^4 & = (a+b+c) (a^3+b^3+c^3-(ab+bc +ca)(a^2+b^2+c^2)+abc(a+b+c) \\ & = 6(5) - 14(8) + \frac{41}{3}(6) \\ & = \boxed{0} \end{aligned}$

Let the $3$ numbers be $a,b,c$ .

By using Newton's Sums we can find a polynomial $P(x$ having $a,b,c$ as its zeroes.

$P(x)=x^3-6x^2+14x-\frac{41}{3}$

By using Newton's Sum we'll get $a^4+b^4+c^4=\boxed{0}$ .