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B
(a+1a−1)(a+1a+1)
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C
(a+1a+1)(a+1a+1)
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D
(a+1a−1)(a+1a−1)
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Solution
The correct option is D(a+1a−1)(a+1a−1) a2+1a2+3−2a−2a =a2+1a2+1+2−2a−2a =(a)2+(1a)2+(−1)2+2×a×1a+2×a×(−1)+2×1a×(−1) ∵(a+b−c)2=a2+b2+c2+2ab−2ac−2cb =(a+1a−1)2=(a+1a−1)(a+1a−1)