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 $a,b,c$?
 A: Note this convenient fact: $$(a+b+c)^3-(a^3+b^3+c^3) = 3(a+b)(b+c)(c+a)$$
Since we know the left-hand side is zero, we also know the right-hand side is zero; this tells us that at least one of these is true:
$$a+b=0 \qquad b+c=0 \qquad c+a=0 \tag{$\star$}$$
We'll go ahead and take the middle one (the others will work out similarly), so that $c = -b$. Then, we have
$$2 = a + b + c = a + 0 = a$$
which gives us $a$. Also
$$6 = a^2 + b^2 + c^2 = 2^2 + b^2 + (-b)^2 = 4 + 2 b^2$$
which gives us $b^2 = 1$, so that $b$ is either $+1$ or $-1$, and $c$ is its opposite.
If we'd made a different choice at $(\star)$, we'd have gotten the same values, $2$, $1$, $-1$, assigned to different variables. That's to be expected, since the original equations are symmetric in $a$, $b$, $c$; the letters are interchangeable. Therefore, the best we can say here is that 

$$\{ a, b, c\} = \{2, 1, -1\}$$

A: 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 $a,b,c$
A: Since the expressions you know the value of are homogeneous and symmetric in $a,b,c$, we know (or should know) they are expressible in terms of homogeneous polynomials in the elementary symmetric functions $E_1=a+b+c$, $E_2=ab+ac+bc$, $E_3=abc$ and in polynomial terms.
In order to find these expressions I prefer a more systematic approach, which can be as follows, rather than plugging in the values and going by attempts.
Degree considerations tell us that
$$
a^2+b^2+c^2=\alpha E_1^2+\beta E_2
$$
With $(a,b,c)=(1,0,0)$ we obtain $1=\alpha$; with $(a,b,c)=(1,1,0)$ we get $2=4\alpha+\beta$. Thus $\beta=-2$.
Also
$$
a^3+b^3+c^3=\alpha E_1^3+\beta E_1E_2+\gamma E_3
$$
With $(a,b,c)=(1,0,0)$ we obtain $1=\alpha$; with $(a,b,c)=(1,1,0)$ we get $2=8\alpha+2\beta=0$; with $(a,b,c)=(1,1,1)$ we get $3=27\alpha+9\beta+\gamma$. Therefore $\beta=-3$ and $\gamma=3$.
The relations are then
$$
\begin{cases}
a^2+b^2+c^2=E_1^2-2E_2\\[4px]
a^3+b^3+c^3=E_1^3-3E_1E_2+3E_3
\end{cases}
$$
Since you're given $E_1=2$, $a^2+b^2+c^2=6$ and $a^3+b^3+c^3=8$, the relations become
$$
E_2=-1,\qquad
E_3=2E_2=-2
$$
Why is this important? Because given $E_1$, $E_2$ and $E_3$, the polynomial having $a,b,c$ as roots is
$$
x^3-E_1x^2+E_2x-E_3
$$
(Viète's formulas). Thus the polynomial is
$$
x^3-2x^2-x+2=x^2(x-2)-(x-2)=(x^2-1)(x-2)=(x-1)(x+2)(x-2)
$$
and you found the roots.
A: $$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(a+c)$$
$$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+ab(2-c)+bc(2-a)+ca(2-b)$$
$$12=8+2(ab+bc+ca)-3abc$$
Now,
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
$$2(ab+bc+ca)=4-6=-2$$
Thus,
$$12=8-2-3abc$$
$$abc=-2$$
Let
$$f(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$
$$f(x)=x^3-2x^2-x+2$$
The values of $a,b,c$ are the roots of $f(x)$.
By trial and error, clearly $1$ is a root.
Then, $$x^3-2x^2-x+2=(x-1)(x^2-x-2)=(x-1)(x+1)(x-2)$$
The values of $a,b,c$ are thus $-1,1,2$ in any order.
A: One possibility to determine this---which is by no means perfect, and will not work all of the time!---is to try a few small integers. The best one to focus on is the second equation:
$$
a^2 + b^2 + c^2 = 6
$$
since all of the terms must be positive. Note that if $a = \pm3$, say, then $a^2 = 9$, which means that all terms have to be from the set $\{-2, -1, 0, 1, 2\}$. They can't all be $\pm1$, since that would be to small, so at least one has to be $\pm2$.
Can you see where to go from here? You can also simplify by noting that the solution is symmetric under swapping $a \to b \to c$ (or any other permutation), so you can assume for simplicity that $a$ is the largest in absolute value.
A: You may use Newton's formulas for recovering the values of $ab+ac+bc$ and $abc$, in order to compute $a,b,c$ from the factorization of $(x-a)(x-b)(x-c)$. As an alternative, we may understand which conditions on $P_1,P_2,P_3$ grant that if $a+b=P_1$ and $a^2+b^2=P_2$, then $a^3+b^3=P_3$. We have:
$$ P_1 P_2 = P_3 + ab(a+b) = P_3+P_1\cdot\frac{P_1^2-P_2}{2} $$
or:
$$ 2 P_3 = 3 P_1 P_2 - P_1^3.$$
Assuming $c=1$, we have $(P_1,P_2,P_3)=(1,5,7)$ that fulfills the previous condition, hence we may assume $a+b=1$ and $a^2+b^2=5$. The solutions are so given by the permutations of $\color{red}{(-1,1,2)}$.
