Mathematics - Determinant Question with Solution | TestHub

Given $\Delta =\left|\begin{array}{ccc}P(a)& P(b)& P(c)\\ a^2& b^2& c^2\\ 1& 1& 1\end{array}\right|$. We have$P(x)=x^3+p{x}^2+qx+r$. Apply the row operation $R_1\to R_1-p{R}_2-r{R}_3$. The elements of the first row become:$P(a)-p{a}^2-r=(a^3+p{a}^2+qa+r)-p{a}^2-r=a^3+qa$$P(b)-p{b}^2-r=(b^3+p{b}^2+qb+r)-p{b}^2-r=b^3+qb$$P(c)-p{c}^2-r=(c^3+p{c}^2+qc+r)-p{c}^2-r=c^3+qc$ So,$\Delta =\left|\begin{array}{ccc}{a}^3+qa& b^3+qb& c^3+qc\\ a^2& b^2& c^2\\ 1& 1& 1\end{array}\right|$. This determinant can be split into two determinants:$\Delta =\left|\begin{array}{ccc}{a}^3& b^3& c^3\\ a^2& b^2& c^2\\ 1& 1& 1\end{array}\right|+\left|\begin{array}{ccc}qa& qb& qc\\ a^2& b^2& c^2\\ 1& 1& 1\end{array}\right|$ The first determinant is a known result:$(a-b)(b-c)(c-a)(ab+bc+ca)$. The second determinant is$q\left|\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ 1& 1& 1\end{array}\right|=q\times (-1)\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=-q(a-b)(b-c)(c-a)$. Combining these,$\Delta =(a-b)(b-c)(c-a)(ab+bc+ca)-q(a-b)(b-c)(c-a)=(a-b)(b-c)(c-a)(ab+bc+ca-q)$.