Mathematics - Determinant Question with Solution | TestHub

The determinant of the coefficient matrix is $D=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)(c-a)$. Using Cramer's Rule:$x=\frac{D_x}{D}=\frac{\left|\begin{array}{ccc}{a}^4& a& a^2\\ b^4& b& b^2\\ c^4& c& c^2\end{array}\right|}{D}=\frac{abc(a-b)(b-c)(c-a)(a+b+c)}{(a-b)(b-c)(c-a)}=abc(a+b+c)$.$y=\frac{D_y}{D}=\frac{\left|\begin{array}{ccc}1& a^4& a^2\\ 1& b^4& b^2\\ 1& c^4& c^2\end{array}\right|}{D}=\frac{(a-b)(b-c)(c-a)(a+b)(b+c)(c+a)}{(a-b)(b-c)(c-a)}=(a+b)(b+c)(c+a)$.$z=\frac{D_z}{D}=\frac{\left|\begin{array}{ccc}1& a& a^4\\ 1& b& b^4\\ 1& c& c^4\end{array}\right|}{D}=\frac{-(a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca)}{(a-b)(b-c)(c-a)}=-(a^2+b^2+c^2+ab+bc+ca)$. We know$a^2+b^2+c^2+ab+bc+ca=(a+b+c{)}^2-2(ab+bc+ca)+(ab+bc+ca)=(a+b+c{)}^2-(ab+bc+ca)$. Therefore,$x+y+z=abc(a+b+c)+(a+b)(b+c)(c+a)-((a+b+c{)}^2-(ab+bc+ca))$.