Question

因式分解：a³+b³+c³-3abc

Answer 1

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

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Answer 2

a³+b³+c³-3abc
=a³+3a²b+3ab²+b³+c³-3a²b-3ab²-3abc
=(a+b)³+c³-3ab(a+b+c)
=(a+b+c)(a²+2ab+b²-ca-cb+c²)-3ab(a+b+c)
=(a+b+c)(a²+b²+c²-ab-ac-bc)