<s>[INST] Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
 [/INST] Here I am giving Solution of yours first question, Next time when you post question,
plz show your try. 
Let $x=a,x=b\;,x=c$ be the roots of the equation $(x-a)(x-b)(x-c) =0$
So $$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0.............(1)$$
Now here $a+b+c=2$ and $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
So we get $ab+bc+ca = -1$ and $$a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]$$
So we get $8-3abc=2[6+1]\Rightarrow abc=-2$
Put all value in equation $(1)\;,$ We get $x^3-2x^2-x+2=0$
So we get $$x^3-2x^2-x+2=(x+1)(x-1)(x-2) =0$$
So we get $x=1\;,-1\;,2$
So we get $$a,b,c\in \left\{-1,1,2\right\}$$, bcz Given equation are symmetrical in $a,b,c$
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