#### Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 41 sub question 2 Maths Textbook Solution.

Answer:$(a-1)^{3}$

Hint Use determinant formula

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

Solution:

$\text { L.H.S }\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|$

\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \& C_{2} \rightarrow C_{2}-C_{3}\\ &=\left|\begin{array}{ccc} a^{2}-1 & 2 a & 1 \\ a-1 & a+1 & 1 \\ 0 & 2 & 1 \end{array}\right|\\ &(a-1) \text { common from } \mathrm{C}_{1}\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 2 a & 1 \\ 1 & a+1 & 1 \\ 0 & 2 & 1 \end{array}\right| \end{aligned}

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