Question

Phân tích a^3+b^3+c^3-3abc thế từ đó suy ra điều kiện của a,b,c để a^3+b^3+c^3=3abc

Answer 1

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

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

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

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

Vậy \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Vậy để \(a^3+b^3+c^3=3abc\) thì \(a+b+c=0\) hoặc \(a=b=c\)