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Question

If a,b and c are real numbers and Δ=∣ ∣b+cc+aa+bc+aa+bb+ca+bb+cc+a∣ ∣=0,
show that either a+b+c=0 or a=b=c.

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Solution

Δ=∣ ∣b+cc+aa+bc+aa+bb+ca+bb+cc+a∣ ∣=0

C1C1+C2+C3
∣ ∣ ∣2(a+b+c)c+aa+b2(a+b+c)a+bb+c2(a+b+c)b+cc+a∣ ∣ ∣=0

Taking 2(a+b+c) common from C1
2(a+b+c)∣ ∣1c+aa+b1a+bb+c1b+cc+a∣ ∣=0

R1R1R2,R2R2R3
2(a+b+c)∣ ∣0cbac0acba1b+cc+a∣ ∣=0

On expanding along first column, we get
2(a+b+c)[(cb)(ba)(ac)2]=0
2(a+b+c)(a2+b2+c2abbcca)=0
(a+b+c)(2a2+2b2+2c22ab2bc2ca)=0
(a+b+c)[(ab)2+(bc)2+(ca)2]=0
(a+b+c)=0 or (ab)2+(bc)2+(ca)2=0
(ab)2+(bc)2+(ca)2=0 , which is possible only when a=b=c
Hence, Δ=0
a+b+c=0 or a=b=c

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