Question

Show that the eliminant of ax + cy + bz = 0, cx + by + az = 0, bx + ay + cz = 0, is $a^3+b^3+c^3-3abc=0.$

Answer 1

$ax+cy+bz=\mathrm{0....}\left(i\right)cx+by+az=\mathrm{0....}\left(ii\right)bx+ay+cz=\mathrm{0....}\left(ii\right)$

Applying cross multiplication on $\left(i\right)$ and $\left(ii\right)$

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

substituing $x,y$ and $z$ in $\left(iii\right)$

$bk(ac-b^2)+ak(bc-a^2)+ck(ab-c^2)=0abc-b^3+abc-a^3+abc-c^3=0a^3+b^3+c^3=3abc$

Hence proved.