Question

Cho \(a^3+b^3+c^3=3abc\)

Tính giá trị biểu thức \(A=\frac{a^{2017}}{b^{2017}}+\frac{b^{2017}}{c^{2017}}+\frac{c^{2017}}{a^{2017}}\)

Answer 1

cm \(a^3+b^3+c^3=3abc\)

thì \(\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)

(chuyển vế xét hiệu ) 

Answer 2

TA CÓ: \(a^3+b^3+c^3=3abc\)

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

\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

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

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow a-b=0;c-a=0;b-c=0\Rightarrow a=b=c\)

\(\Rightarrow\frac{a^{2017}}{b^{2017}}+\frac{b^{2017}}{c^{2017}}+\frac{c^{2017}}{a^{2017}}=1+1+1=3\)