#### Explain Solution R.D.Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 45 maths Textbook Solution.

Answer: $2(a-b)(b-c)(c-a)(a+b+c)$

Hint Use determinant formula

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

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{array}\right| \\ &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1} \\ &=\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| \\ &\text { Use } a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right) \\ &=\left|\begin{array}{ccc} a^{3} & 2 & a \\ (b-a)\left(b^{2}+a b+a^{2}\right) & 0 & b-a \\ (c-a)\left(c^{2}+a c+a^{2}\right) & 0 & c-a \end{array}\right| \end{aligned}

\begin{aligned} &=(-1)(-1)\left|\begin{array}{ccc} a^{3} & 2 & a \\ (a-b)\left(b^{2}+a b+a^{2}\right) & 0 & a-b \\ (c-a)\left(c^{2}+a c+a^{2}\right) & 0 & c-a \end{array}\right|\\ &(a-b)(c-a) \text { common from } \mathrm{R}_{2} \& R_{3}\\ &=(a-b)(c-a)\left|\begin{array}{ccc} a^{3} & 2 & a \\ \left(b^{2}+a b+a^{2}\right) & 0 & 1 \\ \left(c^{2}+a c+a^{2}\right) & 0 & 1 \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{C}_{2}\\ &=(a-b)(c-a)\left[2\left(b^{2}+a b+a^{2}-c^{2}-a c-a^{2}\right)\right]\\ &=2(a-b)(c-a)[(b+c)(b-c)+a(b-c)]\\ &=2(a-b)(b-c)(c-a)(a+b+c)\\ &=R . H . S \end{aligned}