<s>[INST] 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 ?
 [/INST] $$a^2x+a^2y+a^2=0$$
$$\implies (a^2-b^2)y+(a^2-c^2)=0$$
$$a^4x+a^4y+a^4=0$$
$$\implies (a^4-b^4)y+(a^4-c^4)=0$$
If $|a|\neq |b|$ and $|a|\neq |c|$, then ${a^4-b^4\over a^2-b^2}={a^4-c^4\over a^2-c^2}$ which imlies $a^2+b^2=a^2+c^2$ which is $|b|=|c|$
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