Question

Let a,b,c are the solution of the cubic $x^3-5x^2+3x-1=0$, then find the value of the determinant $\left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ b+c& c+a& a+b\end{array}\right|$.

Answer 1

$D=\left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ b+c& c+a& a+b\end{array}\right|$

$x^3-5x^2+3x-1=0$

$a+b+c=5$

$ab+bc+ca=3$

$abc=1$

$R_2\to R_2-R_1$

$=\left|\begin{array}{ccc}a& b& c\\ -b& -c& -a\\ b+c& c+a& a+b\end{array}\right|$

$R_3\to R_3+R_2$

$\left|\begin{array}{ccc}a& b& c\\ -b& -c& -a\\ c& a& b\end{array}\right|=(-1)\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$

$=-\left[a\right(bc-a^2)-b(b^2-ac)+c(ab-c^2\left)\right]$

$=-(abc-a^3-b^3+abc+abc-c^3)$

$=a^3+b^3+c^3-3abc$

$=(a+b+c)[a^2+b^2+c^2-ab-bc-ac]$

$=5(a^2+b^2+c^2-3)$

$=5(31-3)$

$=5\left(28\right)$

$=140$

$(a+b+c{)}^2$

$=a^2+b^2+c^2-2(ab+bc+ac)$

$5^2=a^2+b^2+c^2-6$

$a^2+b^2+c^2=31$.