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Question

# If A=âˆ£âˆ£ âˆ£âˆ£ac2cababbcacaâˆ£âˆ£ âˆ£âˆ£+âˆ£âˆ£ âˆ£âˆ£aabbcbbcccbaâˆ£âˆ£ âˆ£âˆ£+âˆ£âˆ£ âˆ£ âˆ£âˆ£b2cbb+c2a+bcac+ccbaâˆ£âˆ£ âˆ£ âˆ£âˆ£, 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∣∣ ∣ ∣∣ D2→R↔C, D3→R↔C =∣∣ ∣∣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∣∣ ∣∣ C2→C2−cC3 =∣∣ ∣∣a+b2bcab+cabbc+bca∣∣ ∣∣ A=a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)=(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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