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#### Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 2 sub question 8 Maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|=2 x^{3} y^{3} z^{3}$

Hint:We will take common from $C_{1},C_{2} \: \: and\: \: \: C_{3}$

Given:$\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|$

Solution: $\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|$

\begin{aligned} &\text { On taking common } x^{2} \text { from } \mathrm{C}_{1}, \mathrm{y}^{2} \text { from } \mathrm{C}_{2} \text { and } \mathrm{z}^{2} \text { from } \mathrm{C}_{3}\\ &=x^{2} y^{2} z^{2}\left|\begin{array}{ccc} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=x^{2} y^{2} z^{2}\left[0\left|\begin{array}{ll} 0 & y \\ z & 0 \end{array}\right|-x\left|\begin{array}{ll} y & y \\ z & 0 \end{array}\right|+x\left|\begin{array}{ll} y & 0 \\ z & z \end{array}\right|\right]\\ &=x^{2} y^{2} z^{2}[0-x(0 x y-y x z)+x(y x z-0 x z)]\\ &=x^{2} y^{2} z^{2}[-x(0-y z)+x(y z-0)]\\ &=x^{2} y^{2} z^{2}(x y z+x y z)\\ &=x^{2} y^{2} z^{2} \times 2 x y z\\ &=2 x^{3} y^{3} z^{3} \end{aligned}

Hence$\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|=2 x^{3} y^{3} z^{3}$