Question

Cho a + b+ c=0

Tính A= a^3+b^3+c^3-3ab

Answer 1

Sửa đề : Tính \(A=a^3+b^3+c^3-3abc\)

Giải

Ta có : \(A=a^3+b^3+c^3-3abc\)

\(\Leftrightarrow A=\left(a^3+b^3+3a^2b+3b^2a\right)+c^3-3a^2b-3b^2a-3abc\)

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

\(\Leftrightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(\Leftrightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]\)

\(\Leftrightarrow A=0\left(a+b+c=0\right)\)

Vậy \(A=0\)