Question

Giải hệ PT:\(\hept{\begin{cases}x^3-3x=y^3-3y\\x^6+y^6=1\end{cases}}\)

Answer 1

\(\Leftrightarrow\hept{\begin{cases}x^3-y^3-3x+3y=0\\x^6+y^6=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)=0\\x^6+y^6=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)\left(x^2+xy+y^2-3\right)=0\\x^6+y^6=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-y=0\\x^2+xy+y^2-3=0\\x^6+y^6=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=y\\y^2+y.y+y^2-3=0\\y^6+y^6=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=1\\y\approx0,71\end{cases}}\)