System of Three Equations - Prove that at least two numbers' absolute value is equal. Question :

Prove that at least two of the numbers $a,b,c$ must have the absolute values equal in order that the system of following equations in $x,y$ may be consistent
$$ a^2x+b^2y+c^2=0$$
$$a^4x+b^4y+c^4 =0$$
$$x+y+1=0$$

I found $x=\frac{c^2(b^2-c^2)}{a^2(a^2-b^2)}$ and $y=\frac{c^2(c^2-a^2)}{b^2(a^2-b^2)}$ from first two equation.
After putting $x$ and $y$ in $3rd$ equation.
$$\frac{c^2(b^2-c^2)}{a^2(a^2-b^2)}+\frac{c^2(c^2-a^2)}{b^2(a^2-b^2)}+1=0 $$
I did prove it but I am not satisfied with the process as I have to use an identity(never seen and found in the simplification)$$a^2b^2(a^2-b^2)+b^2c^2(b^2-c^2)+c^2a^2(c^2-a^2)=-(a^2-b^2)(b^2-c^2)(c^2-a^2)$$ and the simplification nearly took 30 minutes(which is not good in exam)
Is there any other fast and efficient way to prove or simplify this ?

As $$\begin{pmatrix}1&1&1\\a^2&b^2&c^2\\a^4&b^4&c^4\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix},$$
the matrix must be singular, i.e., a non-trivial linear combinations of the rows is zero, i.e., there are numbers $\alpha,\beta,\gamma$, not all $=0$, such that the polynomial $$
\alpha+\beta t+\gamma t^2$$
has at least three roots $a^2,b^2,c^2$. But such a quadratic (or perhaps even linear) polynomial cannot have more than two distinct roots.
Hence two of these squares must be equal, hence two of $a,b,c$ differ by at most a factor of $-1$ (and therefore have the same absolute value).
