Question

Cho a + b + c = 0. CMR: \(a^3+b^3+c^3=3abc\)

Answer 1

Ta có:

\(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

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

\(=a^3+b^3+3ab\left(a+b\right)\)

\(\Rightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

Thay vào \(a^3+b^3+c^3=0\), ta được:

\(VT=a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3\)

Vì \(a+b+c=0\)

\(\Rightarrow a+b=-c\)

\(\Leftrightarrow a^3+b^3+c^3=\left(-c\right)^{^3}-3ab\left(-c\right)+c^3\)

\(\Leftrightarrow a^3+b^3+c^3=\left(-c\right)^3+c^3+3abc\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

\(\RightarrowĐPCM\).

Answer 2

Có a+b+c=0

=> a+b=-c

=> (a+b)³=-c³

=> a³+b³+3ab(a+b)=-c³

=> a³+b³+3ab(-c)=-c³

=> a³+b³-3abc=-c³

=> a³+b³+c³=3abc