show that [a+b,b+c,c+a]=[abc]2
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Step-by-step explanation:
L.H.S = [(a+b)+(b+c) , (c+a)]
= (a+b) * [ (b+c) × (c+a) ]
= (a+b)* [ b× c + b×a + c× c+ c×a ]
°•° (c × c ) = 0
= a * (b× c ) + b(b× c) + a(b×a) + b(b×a)
= [a b c ] + [a b a ] + [a c a] + [b b c ] - [ b b a ] + [ b c a ]
= [ a b c ] + [ a b c ] = 2[ a b c ] = RHS PROVED
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