#### Explain Solution R.D.Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 34 maths Textbook Solution.

Answer:$2 a^{3} b^{3} c^{3}$

Hint Use determinant formula

Given: $\left|\begin{array}{ccc} 0 & a b^{2} & a c^{2} \\ a^{2} b & 0 & b c^{2} \\ a^{2} c & b^{2} c & 0 \end{array}\right|=2 a^{3} b^{3} c^{3}$

Solution:

$\text { L.H.S }\left|\begin{array}{ccc} 0 & a b^{2} & a c^{2} \\ a^{2} b & 0 & b c^{2} \\ a^{2} c & b^{2} c & 0 \end{array}\right|$

\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &|\Delta|=\left|\begin{array}{ccc} a^{2}(b+c) & b^{2}(a+c) & c^{2}(a+b) \\ a^{2} b & 0 & b c^{2} \\ a^{2} c & b^{2} c & 0 \end{array}\right|\\ &\text { Taking out } \mathrm{a}^{2} \text { from } \mathrm{C}_{1}, b^{2} \text { from } \mathrm{C}_{2} \text { and } \mathrm{c}^{2} \text { from } \mathrm{C}_{3}\\ &=a^{2} b^{2} c^{2}\left|\begin{array}{ccc} b+c & a+c & a+b \\ b & 0 & b \\ c & c & 0 \end{array}\right| \end{aligned}

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