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1
Answer:
ok
Step-by-step explanation:
(a−b)(a+b)+(b−c)(b+c)+(c−a)(c+a)=0
LHS=a
2
−b
2
−ab+ab+b
2
−c
2
+bc−bc+c
2
−a
2
−ac+ac
we see the terms cancel each other hence equation reduced to zero
Or we can also use identity (a−b)(a+b)=a
2
−b
2
to find similar result.
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