Solve one system of equations Solve the system of equations
$$ ax + by + cz = 0, $$
$$ bcx + cay + abz = 0, $$
$$ xyz + abc (a^3x + b^3y + c^3z) = 0 $$
I tried solving this using cross multiplication method  but got stuck at one point :
$$x/ab^2-ac^2 = y/bc^2-ba^2 = z/ca^2-cb^2 = k (say) $$
I substituted the values in the third equation, and after simplifying ended up being here :
$$ k(abc)[a^2c^2(b^2 + a^2c^2 - b^2c^2 - a^2)] $$

As @Aderinsola Joshua commented, use successive eliminations
From $(1)$, you have $y=-\frac{a x+c z}{b}$.
Plug in $(2)$ and then $z=\frac{c x \left(a^2-b^2\right)}{a \left(b^2-c^2\right)}$
Plug in $(3)$ and simplify to get
$$\frac{b c  (a-b) (a+b) (a-c) (a+c) \left(a^2
   \left(b^2-c^2\right)^2-x^2\right)x}{a^2 \left(b^2-c^2\right)^2}=0$$
So, assuming $a\neq b \neq c$, you are left with
$$\left(a^2
   \left(b^2-c^2\right)^2-x^2\right)x=0$$
