Question

Solve the equations :
$ax+by+cz=0,bcx+cay+abz=0,$

$xyz+abc(a^{3}x+b^{3}y+c^{3}z)=0.$

Answer 1

Given, $ax+by+cz=0\to \left(1\right)$

$bcx+cay+abz=0\to \left(2\right)$

$xyz+abc(a^{3}x+b^{3}y+c^{3}z)=0\to \left(3\right)$

$\left(1\right)\times bc-\left(2\right)\times a\Rightarrow abcx+b^{2}cy+b{c}^{2}z-abcx-c{a}^{2}y-a^2{b}^2=0$

$c(b^2-a^2)y=b(a^2-c^2)z$

$\left(1\right)\times ac-\left(2\right)\times b\Rightarrow a^{2}cx+abcy+a{c}^{2}z-b^{2}cx-abcy-a{b}^{2}z=0$

$c(a^2-b^2)x=a(b^2-c^2)z$

$\left(1\right)\times ab-\left(2\right)\times c\Rightarrow a^{2}bx+a{b}^{2}y+abcz-b{c}^{2}x-a{c}^{2}y-abcz=0$

$b(a^2-c^2)x=a(c^2-b^2)y$

Let $c(b^2-a^2)y=b(a^2-c^2)z=k$

$y=\frac{k}{c(b^2-a^2)},z=\frac{k}{b(a^2-c^2)}$

Let $c(a^2-b^2)x=a(b^2-c^2)z=k_1$

$x=\frac{k_1}{c(a^2-b^2)},z=\frac{k_1}{a(b^2-c^{2}0}$

$\frac{k}{b(a^2-c^2)}=\frac{k_1}{a(b^2-c^2)}$

$k_1=\frac{a}{b}\left(\frac{b^2-c^2}{a^2-c^2}\right)k$

$x=\frac{a\left(c^2\right)k}{bc(a^2-b^2)(a^2-c^2)},y=\frac{k}{c(b^2-a^2)},z=\frac{k}{b(a^2-c^2)}$

$xyz=\frac{a(c^2-b^2)k^3}{\left[bc\right(b^2-a^2\left)\right(a^2-c^2){]}^2}$

$a^{3}x+b^{3}y+c^{3}z=a^2(-by-cz)+b^{3}y+c^{3}z$

$=-b(a^2-b^2)y+c(c^2-a^2)z$

$\frac{kb}{c}-\frac{kc}{b}$

$\left(3\right)\Rightarrow xyz+abc(a^{3}x+b^{3}y+c^{3}z)=0$

$\frac{a(c^2-b^2)k^3}{\left(bc\right(c^2-a^2\left)\right(a^2-b^2){)}^2}+ka{b}^2-ka{c}^2=0$

$a(c^2-b^2)k[\frac{k^2}{\left(bc\right(c^2-a^2\left)\right(a^2-b^2){)}^2}-1]=0$

$\therefore k=0$ or $k=\pm bc(c^2-a^2)(a^2-b^2)$

$\therefore$ solutions are $x=0,y=0,z=0$ and

$x=a(b^2-c^2),y=b(a^2-c^2),z=c(a^2-b^2)$

and $x=-a(b^2-c^2),y=-b(a^2-c^2),z=-c(a^2-b^2)$