Question

Cho \(a+b+c=2019\) .

CMR: \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}=2019\)

Answer 1

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

\(\Rightarrow\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-ac-bc}=2019\left(đpcm\right)\)

Answer 2

Ta có : \(\left(a+b+c\right)\left(a^2+b^2+^2-ab-ac-bc\right)\)

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

\(\Leftrightarrow\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-ac-bc}=2019\)

\(\Rightarrowđpcm\)