Solve the equation $x^{4}-3 x^{3}+4 x^{2}-3 x+1=0$.

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Let $y=x+1 / x$. We have $0=\frac{x^{4}-3 x^{3}+4 x^{2}-3 x+1}{x^{2}}=$ $\left(x^{2}+\frac{1}{x^{2}}\right)-3\left(x+\frac{1}{x}\right)+4=\left(y^{2}-2\right)-3 y+4=y^{2}-3 y+2$.

The roots of the polynomial $y^{2}-3 y+2$ are 1 and 2, so $y^{2}-3 y+2=(y-1)(y-2)$. Therefore, the equation is equivalent to $y=1$ or $y=2$.

If $y=2$ then $x+1 / x=2$, so $x^{2}+1=2 x$, which can be rewritten as $(x-1)^{2}=0$. This is equivalent to $x=1$.

If $y=1$ then $x+1 / x=1$, so $x^{2}-x+1=0$. However, $x^{2}-x+1=(x-1 / 2)^{2}+3 / 4>0$, so the equation has no solution.

Conclusion: the unique solution is $x=1$.