<s>[INST] Finding the value of $a+2b+3c$ where $a,b,c$ are roots of a cubic equation. Let $a$, $b$, and $c$ be positive real numbers with $a<b<c$ such that $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=216$. Find $a+2b+3c$.
I solved it like this:
$$2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)$$$$ab+bc+ca=47$$
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$$$abc=60$$
S0 $a,b,c$ are solutions of $x^3-12x^2+47x-60=0$
Placing $x=y+4$$$y^3+12y^2+48y+64-12y^2-96y-192+47y+188-60=0$$$$y^3-y=0$$
So $y=-1,0,1$ and $x=3,4,5$.
Thus $a=3;b=4;c=5$ and $a+2b+3c=26$ 
But is there any other way of solving it without actually making a cubic equation and then solving it to get the values of $a,b,c$
 [/INST] Make the change: $a=x+4, b=y+4,c=z+4$. Then the equations will be:
$$\begin{cases} x+y+z=0 \\
(x+4)^2+(y+4)^2+(z+4)^2=50 \\
(x+4)^3+(y+4)^3+(z+4)^3=216 \end{cases} \Rightarrow 
\begin{cases} x+y+z=0 \\
x^2+y^2+z^2=2 \\
x^3+y^3+z^3=0 \end{cases}$$
Express $x+y=-z$. Square it and subtract the second. Cube it and subtract the third.
$$\begin{cases}xy=-1 \\
3xy(x+y)=0\end{cases} \Rightarrow (x,y,z)=(-1,1,0); (1,-1,0).$$
Note: All $6$ possible solutions will be the permutations of $(-1,1,0)$.
Since $a<b<c$, from the first we get:
$$(a,b,c)=(3,4,5).$$
Note: All $6$ possible solutions without constraint $a<b<c$ will be the permutations of $(3,4,5)$.
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