proove that (a+b+c)^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a)
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(a+b+c)^3-a^3-b^3-c^3
=(a^3+b^3+c^3)+3(a+b)(b+c)(c+a)-(a^3+b^3+c^3)
So,(a^3+b^3+c^3) and -(a^3+b^3+c^3) gets cancelled
Therefore
3(a+b)(b+c)(c+a)=3(a+b)(b+c)(c+a)
=(a^3+b^3+c^3)+3(a+b)(b+c)(c+a)-(a^3+b^3+c^3)
So,(a^3+b^3+c^3) and -(a^3+b^3+c^3) gets cancelled
Therefore
3(a+b)(b+c)(c+a)=3(a+b)(b+c)(c+a)
s55555a:
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