#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 2 Sub Question 6 Maths Textbook Solution.

Answer:$\left|\begin{array}{lll} 1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b \end{array}\right|=0$

Hint: First we will split the determinant into two determinant

Given: $\left|\begin{array}{lll} 1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b \end{array}\right|$

Solution:$\left|\begin{array}{lll} 1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b \end{array}\right|$

\begin{aligned} &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\left|\begin{array}{ccc} 1 & a & b c \\ 1 & b & a c \\ 1 & c & a b \end{array}\right| \\ &\because \text { Using }\left|\begin{array}{lll} x_{1} & x_{2} & x_{3}+a \\ y_{1} & y_{2} & y_{3}+b \\ z_{1} & z_{2} & z_{3}+c \end{array}\right|=\left|\begin{array}{lll} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array}\right|+\left|\begin{array}{lll} x_{1} & x_{2} & a \\ y_{1} & y_{2} & b \\ z_{1} & z_{2} & c \end{array}\right| \\ &==\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & a c \\ 1 & c & a b \end{array}\right| \end{aligned}

We will multiply and divide by abc in second determinent

$=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & a c \\ 1 & c & a b \end{array}\right|$

\begin{aligned} &\text { On multiplying } R_{1} \text { by } a, R_{2} \text { by } b \text { and } R_{3} \text { by } c \text { in determinant }(2)\\ &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\frac{1}{a b c}\left|\begin{array}{lll} a & a^{2} & a b c \\ b & b^{2} & a b c \\ c & c^{2} & a b c \end{array}\right|\\ &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|\\ &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right| \end{aligned}

\begin{aligned} &C_{1} \leftrightarrow C_{2} \text { in determinant }(1)\\ &=-\left|\begin{array}{lll} a & 1 & a^{2} \\ b & 1 & b^{2} \\ c & 1 & c^{2} \end{array}\right|-\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right| \text { and } \end{aligned}

If any two rows or columns of a determinant are interchanged,

then sign of the determinant is changed.

\begin{aligned} &\text { Again } C_{2} \leftrightarrow C_{3} \text { in determinant }(1)\\ &=(-)(-)\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|-\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|\\ &=\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|-\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|\\ &=0 \end{aligned}

Hence$\left|\begin{array}{lll} 1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b \end{array}\right|=0$