Question

Without expanding the determinants, prove that $\left|\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right|=\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|$

Answer 1

$\left|\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right|=\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|$

L.H.S=

$\left|\begin{array}{ccc}a& a^2& bc\\ b& b^2& ca\\ c& c^2& ab\end{array}\right|$

$R_1\leftrightarrow R_1-R_2,R_2\leftrightarrow R_2-R_3$

$=\left|\begin{array}{ccc}(a-b& a^2-b^2& bc-ca\\ (b-c)& b^2-c^2& ca-ab\\ c& c^2& ab\end{array}\right|$$=\left|\begin{array}{ccc}a-b& (a-b)(a+b)& -c(a-b)\\ b-c& (b-c)(b+c)& -a(b-c)\\ c& c^2& ab\end{array}\right|$

$=(a-b)(b-c)\left|\begin{array}{ccc}1& a+b& -c\\ 1& b+c& -a\\ c& c^2& ab\end{array}\right|$

$R_1\leftrightarrow R_1-R_2$

$=(a-b)(b-c)\left|\begin{array}{ccc}0& (a+b)(b+c)& -c+a\\ 1& b+c& -a\\ c& c^2& ab\end{array}\right|$

$=(a-b)(b-c)\left|\begin{array}{ccc}0& a-c& a-c\\ 1& b+c& -a\\ c& c^2& ab\end{array}\right|$

$=(a-b)(b-c)(a-c)\left|\begin{array}{ccc}0& 1& 1\\ 1& b+c& -a\\ c& c^2& ab\end{array}\right|$

$C_3\leftrightarrow C_3-C_2$

$=(a-b)(b-c)(a-c)\left|\begin{array}{ccc}0& 1& 0\\ 1& b+c& -a-b-c\\ c& c^2& ab-c^2\end{array}\right|$

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

$=+(a-b)(b-c)(c-a)(ab+bc+ca)$

$\therefore L.H.S=R.H.S.$ Hence proved.