If a,b,c are the roots of x^3-5x^2+2x-1=0, then the value of beginvmatrixa&b&c\b&c&a\c&a&bendvmatrix is

Expanding the circulant determinant gives $\Delta=3abc-a^3-b^3-c^3=-(a^3+b^3+c^3-3abc)$. Use the identity $a^3+b^3+c^3-3abc=(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)$. From Vieta, $a+b+c=5$ and $ab+bc+ca=2$. Then $a^2+b^2+c^2=25-2\cdot2=21$, so the second factor is $21-2=19$. Therefore $a^3+b^3+c^3-3abc=5\cdot19=95$ and $\Delta=-95$. The sign matters: the circulant determinant is the negative of the usual symmetric expression, so $+95$ is a trap. Cross-check using the factorisation $\Delta=-(a+b+c)\cdot\tfrac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$: the bracket equals $2(21-2)=38$, so $\Delta=-5\cdot19=-95$, which agrees with the value found above.