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Question

If A=∣ ∣ac2cababbcaca∣ ∣+∣ ∣aabbcbbcccba∣ ∣+∣ ∣ ∣b2cbb+c2a+bcac+ccba∣ ∣ ∣,
then A is divisible by

A
a+b+c
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B
a2+b2+c2
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C
a+bw+cw2, where w be the cube root of unity
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D
a+bw2+cw, where w be the cube root of unity
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Solution

The correct option is D a+bw2+cw, where w be the cube root of unity
A=∣ ∣ac2cababbcaca∣ ∣+∣ ∣aabbcbbcccba∣ ∣+∣ ∣ ∣b2cbb+c2a+bcac+ccba∣ ∣ ∣

D2RC, D3RC
=∣ ∣ac2cababbcaca∣ ∣+∣ ∣abcabbcbbcca∣ ∣+∣ ∣b2b+c2cca+bcbbac+ca∣ ∣

Add D1 and D2 as C1, C3 are identical.
=∣ ∣ab+c2caba+bcbbcac+ca∣ ∣+∣ ∣b2b+c2cca+bcbbac+ca∣ ∣

=∣ ∣a+b2b+c2cab+ca+bcbbc+bac+ca∣ ∣

C2C2cC3

=∣ ∣a+b2bcab+cabbc+bca∣ ∣

A=a3+b3+c33abc=(a+b+c)(a2+b2+c2abbcca)=(a+b+c)(a2+w3b2+w3c2+(w+w2)ab+(w+w2)bc+(w+w2)ca)=(a+b+c)(a+bw+cw2)(a+bw2+cw)

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