# Thread: Fantastic system

1. ## Fantastic system

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

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

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

So we have

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

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

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

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

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

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

So we have

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

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

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

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