Question

[Đỗ Thị Kim Tiến]

 Chứng minh: (a+b+c) . (a^2 +b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc

Answer 1

[Yen Nhi]

Chứng minh rằng: \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=a^3+b^3+c^3-3abc\)

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

\(\Rightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) (1)

Thay (1) vào ta được

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

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

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

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

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

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)