Easy , isn't it?

Use Newton's Sums.

Let $a_3 = 1$

$S_1 + a_2 = 0 \Rightarrow a_2 = -16$

$S_2 + a_2S_1 + 2a_1 = 0 \Rightarrow a_1 = 83$

$S_3 + a_2S_2 + a_1S_1 + 3a_0 = 0 \Rightarrow a_0 = -140$

$S_4 + a_2S_3 + a_1S_2 + a_0S_1 = 0 \Rightarrow S_4 = \boxed{3282}$

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Note:

• $S_n = a^n + b^n + c^n$

$(a+b+c)^2$ = $(a^2+b^2+c^2)+2 \times (ab+bc+ca)$ .

Let us substitute $a+b+c=16$ and $a^2+b^2+c^2=90$ in the following equation and after that we get:

$ab+bc+ca=83... (*)$

$(a+b+c)\times (a^2+b^2+c^2)=(a^3+b^3+c^3)+(ca^2+cb^2+c^2a+c^2b+a^2b+b^2a)$

After substituting the values we get: $ca^2+cb^2+c^2a+c^2b+a^2b+b^2a=908...(**)$

$(a+b+c)\times (a^2+b^2+c^2)=(a^3+b^3+c^3)+3\times (ca^2+cb^2+c^2a+c^2b+a^2b+b^2a)+6abc$ .

After substituting the values including ** we get: $abc=140...(***)$

$(a+b+c)\times (a^3+b^3+c^3)=(a^4+b^4+c^4)+ab\times (a^2+b^2)+ca\times (a^2+c^2)+bc\times (b^2+c^2)=$ $(a^4+b^4+c^4)+ab\times (90-c^2)+ca\times (90-b^2)+bc\times (90-a^2)$ $=(a^4+b^4+c^4)+90\times (ab+bc+ca)-abc\times (a+b+c)$

Let us substitute * and * * and the values given in the question in the equation above and we're done: $a^4+b^4+c^4=3282$ .

We can solve the problem using Newton's sums method. Let $P_n = a^n+b^n+c^n$ , $S_1 = a+b+c = 16$ , $S_2 = ab+bc+ca$ and $S_3 = abc$ . Then, we have:

$\begin{aligned} P_2 & = P_1S_1 - 2S2 \\ a^2 + b^2 + c^2 & = (a+b+c)(a+b+c) - 2S_2 \\ 90 & = 16^2 - 2S_2 \\ \implies S_2 & = 83 \\ P3 & = P_2S_1 - P_1S_2 + 3S_3 \\ a^3 + b^3 + c^3 & = (90)(16) - (16)(83) + 3S_3 \\ 532 & = 112 + 3S_3 \\ \implies S_3 & = 140 \\ \implies P_4 & = P_3S_1 - P_2S_2 + P_1S_3 \\ a^4 + b^4 + c^4 & = (532)(16) - (90)(83) + (16)(140) \\ & = \boxed{3282} \end{aligned}$