#### Explain Solution R. D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 52 Sub Question 6 maths Textbook Solution.

Answer: $x=b, c,-(b+c)$

Hint: Use determinant formula

Given:$\left|\begin{array}{lll} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3} \end{array}\right|=0$

Solution:

$\text { LHS }\left|\begin{array}{lll} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3} \end{array}\right|$

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

\begin{aligned} &(b-x)(c-x) \text { common from } \mathrm{R}_{2} \& R_{3} \\ &=(b-x)(c-x)\left|\begin{array}{ccc} 1 & x & x^{3} \\ 0 & 1 & \left(b^{2}+b x+x^{2}\right) \\ 0 & 1 & \left(c^{2}+c x+x^{2}\right) \end{array}\right| \end{aligned}

\begin{aligned} &\text { Expand from } \mathrm{C}_{1}\\ &(b-x)(c-x)\left(c^{2}+c x+x^{2}-b^{2}-b x-x^{2}\right)=0\\ &b-x=0, c-x=0, c^{2}-b^{2}-b x+c x=0\\ &x=b, c=x,(c+b)(c-b)-x(c-b)=0 \end{aligned}