Prove the following identities:

LHS

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

$=\left|\begin{array}{ccc}a^{3} & 2 & a \\ b^{3}-a^{3} & 0 & b-a \\ c^{3}-a^{3} & 0 & c-a\end{array}\right| \quad\left[\right.$ Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

$=-(a-b)(c-a)\left|\begin{array}{ccc}a^{3} & 2 & a \\ b^{2}+a^{2}+a b & 0 & 1 \\ c^{2}+a^{2}+a c & 0 & 1\end{array}\right|$      [Taking $(b-a)$ common from $R_{2}$ and $(c-a)$ common from $R_{3}$ ]

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

$\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{3}\right]$

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

$=-(a-b)(c-a)(b-c)(a+b+c)\left|\begin{array}{ccc}a^{3} & 2 & a \\ 1 & 0 & 0 \\ c^{2}+a^{2}+a c & 0 & 1\end{array}\right|$          [Taking $(b-c)(a+b+c)$ common from $R_{2}$ ]

$=-(a-b)(c-a)(b-c)(a+b+c)(-2)$          [Expanding along second column]

$=2(a-b)(c-a)(b-c)(a+b+c)$

$=\mathrm{RHS}$

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