# Math Help - Fantastic system

1. ## Fantastic system

Find all reals numbers solutions of system :
$\left\{ \begin{array}{l}
{\rm a + b + c = 4} \\
{\rm a}^{\rm 2} + b^2 + c^2 = 14 \\
a^3 + b^3 + c^3 = 34 \\
\end{array} \right.
$

2. $a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)\Rightarrow ab+ac+bc=1$

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc\Rightarrow abc=-6$

So we have

$\left\{\begin{array}{ll}a+b+c=4\\ab+ac+bc=1\\abc=-6\end{array}\right.$

Then $a,b,c$ are the roots of the equation $x^3-4x^2+x+6=0$

The equation has the solutions $x_1=-1, \ x_2=2, \ x_3=3$

Then $\left\{\begin{array}{ll}a=-1\\b=2\\c=3\end{array}\right.$ or $\left\{\begin{array}{ll}a=2\\b=-1\\c=3\end{array}\right.$ or.....etc.

3. Originally Posted by red_dog
$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)\Rightarrow ab+ac+bc=1$

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc\Rightarrow abc=-6$

So we have

$\left\{\begin{array}{ll}a+b+c=4\\ab+ac+bc=1\\abc=-6\end{array}\right.$

Then $a,b,c$ are the roots of the equation $x^3-4x^2+x+6=0$

The equation has the solutions $x_1=-1, \ x_2=2, \ x_3=3$

Then $\left\{\begin{array}{ll}a=-1\\b=2\\c=3\end{array}\right.$ or $\left\{\begin{array}{ll}a=2\\b=-1\\c=3\end{array}\right.$ or.....etc.
Thank you
The system has the solutions
$
\begin{array}{l}
\left\{ \begin{array}{l}
a = 2 \\
b = 3 \\
c = - 1 \\
\end{array} \right.\left\{ \begin{array}{l}
a = 3 \\
b = - 1 \\
c = 2 \\
\end{array} \right.\left\{ \begin{array}{l}
a = 3 \\
b = 2 \\
c = - 1 \\
\end{array} \right. \\
\\
\end{array}

$

And :
$\left\{ \begin{array}{l}
a = - 1 \\
b = 2 \\
c = 3 \\
\end{array} \right.\left\{ \begin{array}{l}
a = - 1 \\
b = 3 \\
c = 2 \\
\end{array} \right.\left\{ \begin{array}{l}
a = 2 \\
b = - 1 \\
c = 3 \\
\end{array} \right.$