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B
(a−b4−c3)(a−b2−c3)
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C
(a−b8−c3)(a−b2−c3)
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D
(a−b7−c3)(a−b2−c3)
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Solution
The correct option is A(a−b2−c3)(a−b2−c3) a2+b24+c29−ab+bc3−23ca =a2+−b222+−c232+2×a×−b2+2×−b2×−c3+2×−c3×a Using, (a+b+c)2=a2+b2+c2+2ab+2bc+2ca =(a−b2−c3)(a−b2−c3)