A cute bare basic proof of factorization of a^{3}+b^{3}+c^{3}-3abc by Gopal/Sravan Pithwa

This proof is not from Titu Andreescu. But original attempt by student Gopal/Sravan Pithwa. It had never occured to me a brute force proof would be so easy in this case :_-) Surely we learn a lot from children/students !!!

Prove that a^{3}+b^{3}+c^{3}-3abic = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

Proof:

LHS = a^{3}+b^{3}+c^{3}-3abc = (a^{3}+b^{3})+ c^{3}-3abc

= (a+b)(a^{2}+b^{2}-ab) + c^{3}-3ab

= (a+b)(a^{2}+b^{2}-ab+3ab-3ab) + c^{3}-3abc

= (a+b)(a^{2}+b^{2}+2ab -3ab)+c^{3}-3abc

=(a+b)((a+b)^{2}-3ab)+c^{3}-3abc

= (a+b)^{3}-3ab(a+b)+c^{3}-3abc

= (a+b)^{3}+c^{3}-3ab(a+b)-3abc

=(a+b+c)((a+b)^{2}+c^{2}-c(a+b))-3ab(a+b+c)

=(a+b+c)(a^{2} + b^{2}+c^{2}+2ab-ca-bc) - 3ab(a+b+c)

= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca).

QED. Hence, a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

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