Question

Using properties of determinants, prove the following: $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

Answer 1

LHS $=\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|$
Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$, we get
$=\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-a\\ a^3& b^3-a^3& c^3-a^3\end{array}\right|$
Taking out (b - a), (c - a) common from $C_2$ and $C_3$ respectively, we get
$=(b-a)(c-a)\left|\begin{array}{ccc}1& 0& 0\\ a& 1& 1\\ a^3& b^2+ab+a^2& c^2+ac+a^2\end{array}\right|$
Expanding along $R_1$, we get
$=-(a-b)(c-a)\left[1\right(c^2+ac+a^2-b^2-ab-a^2)0+0]$
= $-(a-b)(c-a)(c^2+ac-b^2-ab)$
$=-(a-b)(c-a)[-(b^2-c^2)-a(b-c\left)\right]$
$=-(a-b)(c-a)\left[\right(b-c\left)\right(-b-c-a\left)\right]$
$=(a-b)(b-c)(c-a)(a+b+c)]$