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Question

Let P1(x)=ax2bxc, P2(x)=bx2cxa, P3(x)=cx2axb be three quadratic polynomials where a,b,c are non-zero real numbers. Suppose there exists a real number α such that P1(α)=P2(α)=P3(α). Prove that a=b=c.

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Solution

We have three relations:
aα2bαc=λ,
bα2cαc=λ,
cα2aαc=λ,
where λ is the common value. Eliminating α2 from these, taking these equations pair-wise,we get three relations:
(cab2)(bca2)=(ba),(abc2)(cab2)=(cb),(bca2)(abc2)=(ac).
Adding these three, we get
(ab+bc+caa2b2c2)(α1)=0.
(Alternatively, multiplying above relations respectively by bc,ca and ab, and adding also leads to this.) Thus either ab+bc+caa2b2c2=0 or α=1. In the first case
0=ab+bc+caa2b2c2=12((ab)2+(bc)2+(ca)2)
shows that a=b=c. If α=1, then we obtain
abc=bca=cab,
a=b=c.

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