Question

Using properties of determinants prove the following questions.

$\left|\begin{array}{ccc}x& x^2& 1+p{x}^3\\ y& y^2& 1+p{y}^3\\ z& z^2& 1+p{z}^3\end{array}\right|=(1+pxyz)(x-y)(y-z)(z-x)$, where p is any scalar.

Answer 1

$LHS=\left|\begin{array}{ccc}x& x^2& 1+p{x}^3\\ y& y^2& 1+p{y}^3\\ z& z^2& 1+p{z}^3\end{array}\right|=\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\left|\begin{array}{ccc}x& x^2& p{x}^3\\ y& y^2& p{y}^3\\ z& z^2& p{z}^3\end{array}\right|$
In the first determinant, interchange $C_1$ and $C_3,C_2$ and $C_3$ and in the second determinant take out p from $C_3$. we get
$=(-1{)}^2\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|+p\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|+pxyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$
[Taking out x from $R_1$, y from $R_2$ and z from $R_3$ in the second determinant]
$=(1+pxyz)\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=1(1+pxyz)\left|\begin{array}{ccc}1& x& x^2\\ 0& y-x& y^2-x^2\\ 0& z-x& z^2-x^2\end{array}\right|$
(using $R_2\to R_2-R_{1}and{R}_3\to R_3-R_1$)
Expanding along $C_1$, we get
$=(1+pxyz)\left[\right(y-x\left)\right(z^2-x^2)-(z-x\left)\right(y^2-x^2\left)\right]=(1+pxyz)\left[\right(y-x\left)\right(z-x\left)\right(z+x)-(z-x\left)\right(y-x\left)\right(y+x\left)\right]=(1+pxyz)(y-x)(z-x)\left[\right(z+x)-(y+x\left)\right]=(1+pxyz)(y-x)(z-x)\left[\right(z+x-y-x\left)\right]=(1+pxyz)\left[\right(y-x\left)\right(z-x\left)\right(z-y\left)\right]=(1+pxyz)(x-y)(y-z)(z-x)$