Question

Using properties of determinants, prove that $\left|\begin{array}{ccc}{a}^3& 2& a\\ b^3& 2& b\\ c^3& 2& c\end{array}\right|=2(a-b)(b-c)(c-a)(a+b+c)$.

Answer 1

LHS : Let $\Delta =\left|\begin{array}{ccc}{a}^3& 2& a\\ b^3& 2& b\\ c^3& 2& c\end{array}\right|$ By $R_1\to R_1-R_2,R_2\to R_2-R_3$

$\Rightarrow \Delta =\left|\begin{array}{ccc}{a}^3-b^3& 0& a-b\\ b^3-c^3& 0& b-c\\ c^3& 2& c\end{array}\right|$ Taking (a-b) & (b-c) common from $R_1$ and $R_2$ respectively

$\Rightarrow \Delta =(a-b)(b-c)\left|\begin{array}{ccc}{a}^2+ab+b^2& 0& 1\\ b^2+bc+c^2& 0& 1\\ c^3& 2& c\end{array}\right|$ By $R_1\to R_1-R_2$

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

Expanding along $R_1$

$\Rightarrow \Delta =(a-b)(b-c)(c-a)\left\{\right(-a-b-c\left)\right[0-2]-0+0\}$

$\therefore \Delta$ =2(a-b)(b-c)(c-a)(a+b+c)= RHS.