\(x^6-7x^3-8=0\)
\(x^6+x^3-8x^3-8=0\)
\(x^3\left(x^3+1\right)-8\left(x^3+1\right)=0\)
\(\left(x^3+1\right)\left(x^3-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^3+1=0\\x^3-8=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy........
\(x^6-7x^3-8=0\)
\(\Rightarrow x^6+x^3-8x^3-8=0\)
\(\Rightarrow x^3\left(x^3+1\right)-8\left(x^3+1\right)=0\)
\(\Rightarrow\left(x^3+1\right)\left(x^3-8\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2-x+1\right)\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-x+1=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\)
Mà ta có:
\(x^2-x+1=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(x^2+2x+4=x^2+2x+1+3\)
\(=\left(x+1\right)^2+3\)
=> \(x^2-x+1\) và \(x^2+2x+4\) đều vô nghiệm
=> \(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)