Question

Using properties of determinant, prove that:

$\left|\begin{array}{ccc}x& x^2& 1+{px}^3\\ y& y^2& 1+{py}^3\\ z& z^2& 1+{pz}^3\end{array}\right|$$=(1+pxyz)(x-y)(y-z)(z-x)$

Answer 1

Consider, $\left|\begin{array}{ccc}x& x^2& 1+{px}^3\\ y& y^2& 1+{py}^3\\ z& z^2& 1+{pz}^3\end{array}\right|$

$R_2\to R_2-R_1,R_3\to R_3-R_1$

$=\left|\begin{array}{ccc}x& x^2& 1+{px}^3\\ y-x& y^2-x^2& {p(y}^3-x^3)\\ z-x& z^2-x^2& {p(z}^3-x^3)\end{array}\right|$

Taking $y-x$ common from $R_2$ and $z-x$ common from $R_3$

$=(y-x)(z-x)\left|\begin{array}{ccc}x& x^2& 1+{px}^3\\ 1& y+x& {p(y}^2+x^2+xy)\\ 1& z+x& {p(z}^2+x^2+zx)\end{array}\right|$

$R_2\to R_2-R_3$

$=(y-x)(z-x)\left|\begin{array}{ccc}x& x^2& 1+{px}^3\\ 0& y-z& {p(y}^2-z^2+xy-zx)\\ 1& z+x& {p(z}^2+x^2+zx)\end{array}\right|$

$=(y-x)(z-x)\left|\begin{array}{ccc}x& x^2& 1+{px}^3\\ 0& y-z& p\left[\right(y-z\left)\right(y+z)+x(y-z\left)\right]\\ 1& z+x& {p(z}^2+x^2+zx)\end{array}\right|$

Taking $y-x$ common from $R_2$

$=(y-x)(z-x)(y-z)\left|\begin{array}{ccc}x& x^2& 1+{px}^3\\ 0& 1& p(x+y+z)\\ 1& z+x& {p(z}^2+x^2+zx)\end{array}\right|$

Expanding along the first column, we get

$(y-x)(z-x)(y-z)[x{\left\{p\right(z}^2+x^2+zx)-p(xz+x^2+yz+xy+zx+z^2)\left\}\right]+1\left[x^{2}p\right(x+y+z)-1-p{x}^3]$

$=(y-x)(z-x)(y-z)[px{z}^2+p{x}^3+pz{x}^2-p{x}^{2}z-p{x}^3-pxyz-p{x}^{2}y-p{x}^{2}z-p{x}^2{z}^2]$

$+[{px}^3+{px}^{2}y+{px}^{2}z-1-p{x}^3]$

$=(y-x)(z-x)(y-z)(-1-pxyz)$

$=(x-y)(y-z)(z-x)(1+pxyz)$