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 :
$$ k(abc)[a^2c^2(b^2 + a^2c^2 - b^2c^2 - a^2)] $$

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.
