Question

solve this question ​

Answer 1

Step-by-step explanation:

$\pmb{\[ \text { Let L.H.S. }=\Delta=\left|\begin{array}{ll} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array}\right| \]}$

$\text{Multiplying \( \tt R_{1}, R_{2} \) and \( \tt R_{3} \) by \( \tt x, y \) and \( \tt z \) respectively,}$

we get

$\pmb{\[ \Delta=\frac{1}{x y z}\left|\begin{array}{ccc} x^{2} & x^{3} & x y z \\ y^{2} & y^{3} & x y z \\ z^{2} & z^{3} & x y z \end{array}\right|} \\ \\ \pmb{ \tt =\frac{x y z}{x y z}\left|\begin{array}{clc} x^{2} & x^{3} & 1 \\ y^{2} & y^{3} & 1 \\ z^{2} & z^{3} & 1 \end{array}\right| \]}$

$\pmb{ \text{[Taking \( x y z \) common from \( C_{3} \) ]}}$

$\text{Applying \( C_{2} \longleftrightarrow C_{3} \)}$

, we get

$\[ -\left|\begin{array}{ccl} x^{2} & 1 & x^{3} \\ y^{2} & 1 & y^{3} \\ z^{2} & 1 & z^{3} \end{array}\right| \]$

$\text{Applying \( C_{1} \longleftrightarrow C_{2} \)}$

, we get

$\[ \left|\begin{array}{ll} 1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3} \end{array}\right| \]$

$\text{Applying \( \tt R_{2} \rightarrow R_{2}-R_{1} \) and \( \sf R_{3} \rightarrow R_{3}-R_{1} \), }$

we get

$\[ \left|\begin{array}{rcr} 1 & x^{2} & x^{3} \\ 0 & y^{2}-x^{2} & y^{3}-x^{3} \\ 0 & z^{2}-x^{2} & z^{3}-x^{3} \end{array}\right| \]$

$\text{Taking \( \sf (y-x) \) and \( \sf(z-x) \) common from \( \sf R_{2} \) and \( \sf R_{3} \) respectively,}$

we get

$\sf \[ (y-x)(z-x)\left|\begin{array}{ccr} 1 & x^{2} & x^{3} \\ 0 & y+x & y^{2}+y x+x^{2} \\ 0 & z+x & z^{2}+z x+x^{2} \end{array}\right| \]$

$\text{Applying \( \tt R_{2} \rightarrow R_{2}-R_{3} \),}$

we get

$\sf\[ (y-x)(z-x)\left|\begin{array}{rrc} 1 & x^{2} & x^{3} \\ 0 & y-z & \: \: \: \: \: \: \: \: y^{2}-z^{2}+x(y-z) \\ 0 & z+x & z^{2}+z x+x^{2} \end{array}\right| \]$

$\text{Taking \( \sf(y-z) \) common from \( \sf R_{2} \),}$

we get

$\pmb{ \sf\[ (y-x)(z-x)(y-z)\left|\begin{array}{rrc} 1 & x^{2} & x^{3} \\ 0 & 1 & x+y+z \\ 0 & z+x & z^{2}+z x+x^{2} \end{array}\right| \]}$

$\text{Expanding along \( C_{1} \),}$

we get

$\color{darkcyan}\[ \begin{array}{l} \bf (y-x)(z-x)(y-z)\left[\left(z^{2}+z x+x^{2}\right)-(z+x)(x+y+z)\right] \\ \\ \bf =(y-x)(z-x)(y-z)\left[z^{2}+z x+x^{2}-z x-z y-z^{2}-x^{2}-x y-z x\right] \\ \\ \bf =(y-x)(z-x)(y-z)[-x y-x z-z y] \\ \\ \bf =(x-y)(y-z)(z-x)(x y+y z+z x)=\text { R.H.S. } \end{array} \]$

Hence, proved.