<s>[INST] 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 :
$$ k(abc)[a^2c^2(b^2 + a^2c^2 - b^2c^2 - a^2)] $$
 [/INST] Let us consider $b \neq c, a \neq b, a \neq c$. Then from the first two equation we get
\begin{align}
\left[\begin{array}{cc}b&c\\c&b\end{array}\right]\left[\begin{array}{c}y\\z\end{array}\right]=\left[\begin{array}{c}-a\\-\frac{bc}{a}\end{array}\right]x,
\end{align}
i.e.
\begin{align}
\left[\begin{array}{c}y\\z\end{array}\right]=\frac{1}{a(b^2-c^2)}\left[\begin{array}{c}b(c^2-a^2)\\c(a^2-b^2)\end{array}\right]x.
\end{align}
By substitution to the third Eq. we get either $x=0$ or
\begin{align}
x^2=\frac{a^2(b^2-c^2)\left[a^4(c^2-b^2)+b^4(a^2-c^2)+c^4(b^2-a^2)\right]}{(c^2-a^2)(a^2-b^2)}\equiv a^2(b^2-c^2)^2.
\end{align}
Let us check the solution. For $a=1,b=2,c=3$ we get solution $x=5,y=-16,z=9$ (and also $x=-5,y=16,z=-9$), which are the correct solutions of the system of equations given above. $\spadesuit$
</s>