`F(x)=x-8x^3`

Đặt `F(x)=0`

`=>x-8x^3=0`

`=>x.(1-8x^2)=0`

`=>`\(\left[ \begin{array}{l}x=0\\1-8x^2=0\end{array} \right.\) 

`=>`\(\left[ \begin{array}{l}x=0\\8x^2=1\end{array} \right.\) 

`=>`\(\left[ \begin{array}{l}x=0\\x^2=\dfrac{1}{8}\end{array} \right.\) 

`=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{2\sqrt{2}}\\x=-\dfrac{1}{2\sqrt{2}}\end{array} \right.\) 

Vậy: `x in {0; +- 1/(2\sqrt{2})}`