Question

Solve the system of equations: $\begin{array}{c}z+ay+a^{2}x+a^3=0\\ z+by+b^{2}x+b^3=0\\ z+cy+c^{2}x+c^3=0\end{array}\}$ where $a\ne b\ne c$

• A. $x=-(a+b+c),y=ab+bc+ca,z=-abc$
• B. $x=(a+b+c),y=ab+bc+ca,z=abc$
• C. No solution
  
• D. None of these
  

Answer 1

The correct option is A $x=-(a+b+c),y=ab+bc+ca,z=-abc$

Augumented matrix is $\left[\begin{array}{cccc}{a}^2& a& 1& -a^3\\ b^2& b& 1& -b^3\\ c^2& c& 1& -c^3\end{array}\right]$
Using crammer's rule
$\Delta =\left|\begin{array}{ccc}{a}^2& a& 1\\ b^2& b& 1\\ c^2& c& 1\end{array}\right|=-\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=-(a-b)(b-c)(c-a)$
${\Delta }_1=\left|\begin{array}{ccc}-a^3& a& 1\\ -b^3& b& 1\\ -c^3& c& 1\end{array}\right|=(a+b+c)(a-b)(b-c)(c-a)$
${\Delta }_2=\left|\begin{array}{ccc}{a}^2& -a^3& 1\\ b^2& -b^3& 1\\ c^2& -c^3& 1\end{array}\right|=-(ab+bc+ca)(a-b)(b-c)(c-a)$
${\Delta }_3=\left|\begin{array}{ccc}{a}^2& a& -a^3\\ b^2& b& -b^3\\ c^2& c& -c^3\end{array}\right|=abc(a-b)(b-c)(c-a)$
$\therefore x=\frac{{\Delta }_1}{\Delta }=-(a+b+c),y=\frac{{\Delta }_2}{\Delta }=ab+bc+ca,z=\frac{{\Delta }_3}{\Delta }=-abc$