Question

Solve the equations
$ax+by+cz=0...\left(1\right),$
$x+y+z=0...\left(2\right),$
$bcx+cay+abz=(b-c)(c-a)(a-b)...\left(3\right)$.

Answer 1

Given equations are
$ax+by+cz=0...\left(1\right),$
$x+y+z=0...\left(2\right),$

$bcx+cay+abz=(b-c)(c-a)(a-b)...\left(3\right)$.

From $\left(1\right)$ and $\left(2\right)$ by cross multiplication,
$\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}=k,$ suppose ;
$\therefore x=k(b-c),y=k(c-a),z=k(a-b).$
Substituting in $\left(3\right),$ we get
$k\left\{bc\right(b-c)+ca(c-a)+ab(a-b\left)\right\}=(b-c)(c-a)(a-c)$
$k\{-(b-c\left)\right(c-a\left)\right(a-b\left)\right\}=(b-c)(c-a)(a-b)$
$\therefore k=-1;$
Hence $x=c-b,y=a-c,z=b-a$.