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B
False
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Solution
The correct option is A True L.H.S:(a+b)3+(b+c)3+(c+a)3−3(a+b)(b+c)(c+a) =(a3+3a2b+3ab2+b3)+(b3+3b2c+3bc2+c3)+(c3+3c2a+3ca2+a3)−3(a+b)(bc+ba+c2+ca) =(2a3+2b3+2c3+3a2b+3ab2+3b2c+3bc2+3c2a+3ca2)−3(a+b)(bc+ba+c2+ca) =(2a3+2b3+2c3+3a2b+3ab2+3b2c+3bc2+3c2a+3ca2)−3(abc+ba2+ac2+ca2+b2c+b2a+bc2+abc) =2a3+2b3+2c3−6abc =2(a3+b3+c3−3abc) =R.H.S