#### Question

Prove a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)

Collected in the board: The сube of sum

Steven Zheng posted 1 year ago

Using the formula of the сube of sum of two expressions, which shows the сube of the first, plus three times the product of the square of the first and the second, plus three times the product of the first and the square of second, plus the сube of the second

(a+b)^3=a^3+3a^2b+3ab^2+b^3

\therefore a^3+b^3=(a+b)^3-3a^2b-3ab^2

\therefore a^3+b^3+c^3-3abc

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

=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)

=(a+b+c)(a^2+2ab+b^2+c^2-ac-bc-3ab)

=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)

Steven Zheng posted 1 year ago

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