Question

Suppose $a,b,c$ are distinct and $x,y,z$ are connected by the system of equations
$x+ay+a^{2}z=a^3$
$x+by+b^{2}z=b^3$
$x+cy+c^{2}z=c^3$
then

Answer 1

$x+ay+a^{2}z=a^{3}x+by+b^{2}z=b^{3}x+cy+c^{2}z=c^3$
$M=\left[\begin{array}{cc}\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}& \begin{array}{c}{a}^3\\ b^3\\ c^3\end{array}\end{array}\right]$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
$M=\left[\begin{array}{cc}\begin{array}{ccc}1& a& a^2\\ 0& b-a& b^2-a^2\\ 0& c-a& c^2-a^2\end{array}& \begin{array}{c}{a}^3\\ b^3-a^3\\ c^3-a^3\end{array}\end{array}\right]$
Now dividing $R_2$ by $b-a$ and $R_3$ by $c-a$
$M=\left[\begin{array}{cc}\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 1& c+a\end{array}& \begin{array}{c}{a}^3\\ b^2+a^2+ab\\ c^2+a^2+ac\end{array}\end{array}\right]$
Applying $R_3\to R_3-R_2$
$M=\left[\begin{array}{cc}\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 0& c-b\end{array}& \begin{array}{c}{a}^3\\ b^2+a^2+ab\\ c^2-b^2+a(c-b)\end{array}\end{array}\right]$
On solving we get
$z=a+b+c,y=-(ab+bc+ac),x=abc$
A) $a+b+c=z$
B) $a+b+c=-y$
C) $abc=x$
D) $(b+c)(c+a)(a+b)=(bc+ba+c^2+ac)(a+b)=abc+a^{2}b+c^{2}a+a^{2}c+b^{2}c+b^{2}a+c^{2}b+abc=-abc+(ab+bc+ac)(a+b+c)=-x-yz$