# Using properties of determinants, prove that: $\begin{vmatrix} x &x^2 &1 & px^3\\ y &y^2 &1 & py^3\\ z &z^2 &1 & pz^3 \end{vmatrix}= (1+pxyz) (x-z) (y-z)(z-x),$ where p is any scalar.

Toolbox:
• If each element of a row(or a column) of a determinant is multiplied by a constant k,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any column of a given determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\Delta = \begin{vmatrix} x & x^2& 1+px^3\\y & y^2 & 1+py^3\\z&z^2&1+pz^3\end{vmatrix}$
Apply $R_2\rightarrow R_2-R_1$ and $R_3\rightarrow R_3-R_1$ we have,
$\Delta = \begin{vmatrix} x & x^2& 1+px^3\\y-x & y^2-x^2 & p(y^3-x^3)\\z-x&z^2-x^2&p(z^3-x^3)\end{vmatrix}$
Taking (y-x) from $R_2$ and (z-x) from $R_3$ as common factors,
$\Delta =(y-z)(z-x) \begin{vmatrix} x & x^2& 1+px^3\\1 & y+x & p(y^2+x^2+xy)\\1&z+x&p(z^2+x^2+x^2)\end{vmatrix}$
Apply $R_3\rightarrow R_3-R_2$ we get,
$\Delta =(y-z)(z-x) \begin{vmatrix} x & x^2& 1+px^3\\1 & y+x & p(y^2+x^2+xy)\\0&z-y&p(z-y)(x+y+z)\end{vmatrix}$
Take (z-y) as a common factor from $R_3$
$\Delta =(y-z)(z-x)(z-y) \begin{vmatrix} x & x^2& 1+px^3\\1 & y+x & p(y^2+x^2+xy)\\0&1&p(x+y+z)\end{vmatrix}$
Now expanding along $R_3$,
$\Delta =(x-y)(y-x)(z-x)[(-1)p(xy^2+x^2+x^2y)+1+px^3+p(x+y+z)(xy)]$
$\Delta =(x-y)(y-x)(z-x)[-pxy^2-px^3-px^2y+1+px^3+px^2y+pxy^2+pxyz]$
$\;\;\; =(x-y)(y-x)(z-x)[1+pxyz]$
Hence proved.

edited Mar 5, 2013