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 $$\left(a^2 \left(b^2-c^2\right)^2-x^2\right)x=0$$
Here is a systematic method that brings back your issue to solving a linear system.
I assume $abc \ne 0$. Setting
$$X=x/a, Y=y/b, Z=z/c \ \ \ \text{and} \ \ \ A=a^2,B=b^2,C=c^2,$$
then replacing $x,y,z$ by $aX,bY,cZ$ resp., and inverting equ. 1, with equ. 2, your system can be written under the form:
$$\begin{pmatrix}1&1&1\\A&B&C\\A^2&B^2&C^2\end{pmatrix}\begin{pmatrix}X\\Y\\Z\end{pmatrix}=\begin{pmatrix}0\\0\\-XYZ\end{pmatrix}$$
Otherwise said:
$$\begin{pmatrix}1&1&1\\A&B&C\\A^2&B^2&C^2\end{pmatrix}\begin{pmatrix}1/(YZ)\\1/(XZ)\\1/(XY)\end{pmatrix}=\begin{pmatrix}0\\0\\-1\end{pmatrix}$$
It remains to solve a simple linear system (with a so-called "Vandermonde matrix", always invertible, because its determinant is $(B-A)(C-A)(C-B) \ne 0$) then obtain $X,Y,Z$ out of the result.
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. $\spadesuit$
$$x = a(c-b)(c+b)$$ $$y = b(a-c)(a+c)$$ $$z = c(b-a)(b+a)$$
