If $3-\sqrt{2}$  is a root of $x^4-8x^3+21x^2-26x+14=0$ then the roots are

since $3-\sqrt{2} is root of the given equation \Rightarrow 3+\sqrt{2} is also root of it.$

$$\begin{gathered} \left(x-\left(3-\sqrt{2}\right)\right)\left(x-\left(3+\sqrt{2}\right)\right) is a factor of it. \\ i.e x^2-6x+7 is factoe of it. \\ Now devide the given equation with x^2-6x+7 \end{gathered}$$

$$\begin{gathered} \therefore x^4-8x^3+21x^2-26x+14=\left(x^2-6x+7\right)\left(x^2-2x+2\right)=0 \\ \Rightarrow x=3\pm \sqrt{2},1\pm i \end{gathered}$$