Linear Quadratic Cubic Quartic

Think of $a, b, c$ as roots of a cubic polynomial, by Vieta's Formula, we know it has the following shape: $p(x)=x^{3}-2x^{2}+\alpha x+\beta$ By using Newton's theorem, we calculate $\alpha$ and $\beta$ : $S_{2}-2S_{1}+\alpha S_{0}+\beta S_{-1}=0$ By using the reciprocal polynomial of $p(x)$ , and Vieta's formulas, we calculate $S_{-1}$ : $p^{-1}(x)=\beta x^{3}+ \alpha x^{2}-2x+1$ $\Rightarrow S_{-1}=\frac{-\alpha}{\beta}$ Therefore: $6-2.2+\alpha.3+\beta . \frac{-\alpha}{\beta}=0$ $\alpha=-1$ Again, we calculate $\beta$ : $S_{3}-2S_{2}-S_{1}+\beta S_{0}=0$ $8-2.6-2+\beta .3=0$ $\beta=2$ The polynomial is then: $p(x)=x^{3}-2x^{2}-x+2$ By using Newton's theorem: $S_{4}-2S_{3}-S_{2}+2S_{1}=0$ $S_{4}-2.8-6+2.2=0$ $S_{4}=18.$

$4=(a+b+c)^2 = a^2+b^2+c^2+2[ ab+bc+ca ]=6+2[ ab+bc+ca ]\\\implies~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~ab+bc+ca = - 1.................(1)\\~~\\8=a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2- [ ab+bc+ca ])+3abc\\=2*(6+1)+3abc~~~\implies~3abc=8-14~~~~~~~~~~~~~~~~abc=-2.............(2)\\From (1)\\1=(ab+bc+ca )^2= a^2b^2+b^2c^2+c^2a^2+2(ab^2c+abc^2+a^2bc)\\=a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)\\=a^2b^2+b^2c^2 +c^2a^2+2(-2)2.~~~~~\therefore a^2b^2+b^2c^2+c^2a^2=9.....(3)\\ ~~\\36=(a^2+b^2+c^2)^2 = a^4+b^4+c^4+2[a^2b^2+b^2c^2+c^2a^2]\\From (3)~~~~~~~~~ 36= a^4+b^4+c^4+2*9~ ~~~~~\implies~a^4+b^4+c^4 = ~~~~\boxed{\color{#D61F06}{\large 1}} \\~~\\$ $\large \text{ For those who are familiar with} \sum_{cyl}~~and~~\prod_{cyl}$

$4=(\sum_{cyl}a)^2=\sum_{cyl}a^2+\sum_{cyl}ab=6+2\sum_{cyl}ab~~\\\implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\sum_{cyl}ab=-1..(1)\\~~\\8=\sum_{cyl}a^3=\sum_{cyl}a*(\sum_{cyl}a^2-\sum_{cyl}ab ) + 3\prod_{cyl}a \\=2*( 6+ 1)+3\prod_{cyl}a~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~~~\prod_{cyl}a= -2...(2)\\~~\\From ~(1)~~1=(\sum_{cyl}ab)^2=\sum_{cyl}a^2b^2+2*\sum_{cyl}a^2bc \\ =\sum_{cyl}a^2b^2+2\prod_{cyl}a*\sum_{cyl}a^2b^2= \sum_{cyl}a^2b^2+2*(-2)*2~~\\\implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\sum_{cyl}a^2b^2=9..(3)\\~~\\From ~(3)~~36=(\sum_{cyl}a^2)^2=\sum_{cyl}a^4+2\sum_{cyl}a^2b^2=\sum_{cyl}a^4+2*9\\\therefore~~\sum_{cyl}a^4= ~~~~~~~~~~~~~~~~~~~~~~~ \boxed{\color{#D61F06}{\large 1}}$