Question

$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^3& c^3\end{array}\right|$
Prove that : = $(a-b)(b-c)(c-a)(a+b+c).$

Answer 1

Now,

$\left|\begin{array}{ccc}1& a& a^3\\ 1& b& b^3\\ 1& c& c^3\end{array}\right|$

[$R_2^{\prime }=R_2-R_1$ and $R_3^{\prime }=R_3-R_1$]

$=\left|\begin{array}{ccc}1& a& a^3\\ 0& b-a& b^3-a^3\\ 0& c-a& c^3-a^3\end{array}\right|$

$=(b-a)(c-a)\left|\begin{array}{ccc}1& a& a^3\\ 0& 1& b^2+ab+a^2\\ 0& 1& c^2+ac+a^2\end{array}\right|$

$=-(b-a)(c-a)(b^2-c^2+a(b-c\left)\right)$

$=-(b-a)(c-a)(b-c)(a+b+c)$

$=(a-b)(b-c)(c-a)(a+b+c)$.