Using properties of determinants, prove the following: `|xx^2 1+p x^3y y^2 1+p y^3z z^2 1+p z^3|=(1+p x y z)(x-y)(y-z)(z-x)`

Explanation

To Prove that`|(x, x^2, 1+p x^3),( y, y^2, 1+p y^3),(z, z^2, 1+p z^3)|=(1+p x y z)(x-y)(y-z)(z-x)`

Solving L.H.S.

`|(x, x^2, 1+p x^3),( y, y^2, 1+p y^3),(z, z^2, 1+p z^3)|`

Expressing elements of `3^(rd)` column as sum of two elements

` " " " " =|(x, x^2, 1),( y, y^2, 1),(z, z^2, 1)|+|(x, x^2, p x^3),( y, y^2, p y^3),(z, z^2, p z^3)|`

Taking p common from `C_3` in `2^(nd)` matrix

` " " " " =|(x, x^2, 1),( y, y^2, 1),(z, z^2, 1)|+p|(x, x^2, x^3),( y, y^2, y^3),(z, z^2, z^3)|`

In matrix `2^(nd)`, Take x common frpm `R_1`, y common from `R_2` and z common from `R_3`

` " " " " =|(x, x^2, 1),( y, y^2, 1),(z, z^2, 1)|+pxyz|(1, x, x^2),( 1 ,y, y^2),(1, z, z^2)|`

In matrix `2^(nd)`, Replace `C_1 leftrightarrow C_2`

` " " " " =|(x, x^2, 1),( y, y^2, 1),(z, z^2, 1)|-pxyz|(x, 1, x^2),( y ,1, y^2),(z, 1, z^2)|`

In matrix `2^(nd)`, Again Replace `C_2 leftrightarrow C_3`

` " " " " =|(x, x^2, 1),( y, y^2, 1),(z, z^2, 1)|-(-1)pxyz|(x, x^2,1 ),( y ,y^2, 1 ),(z, z^2, 1 )|`

` " " " " =|(x, x^2, 1),( y, y^2, 1),(z, z^2, 1)|+pxyz|(x, x^2,1 ),( y ,y^2, 1 ),(z, z^2, 1 )|`

Taking common `|(x, x^2, 1),( y, y^2, 1),(z, z^2, 1)|`

` " " " " =(1+pxyz)|(x, x^2,1 ),( y ,y^2, 1 ),(z, z^2, 1 )|`

Applying `R_1 to R_1-R_2`

` " " " " =(1+pxyz)|(x-y, x^2-y^2,1-1 ),( y ,y^2, 1 ),(z, z^2, 1 )|`

` " " " " =(1+pxyz)|(x-y, (x-y)(x+y),0 ),( y ,y^2, 1 ),(z, z^2, 1 )|`

Taking common `(x-y)` from `R_1`

` " " " " =(1+pxyz)(x-y)|(1,x+y,0 ),( y ,y^2, 1 ),(z, z^2, 1 )|`

Applying `R_2 to R_2-R_3`

` " " " " =(1+pxyz)(x-y)|(1,x+y,0 ),( y-z ,y^2-z^2, 1-1 ),(z, z^2, 1 )|`

` " " " " =(1+pxyz)(x-y)|(1,x+y,0 ),( y-z ,(y-z)(y+z), 0 ),(z, z^2, 1 )|`

Taking common `(y-z)` from `R_2`

` " " " " =(1+pxyz)(x-y)(y-z)|(1,x+y,0 ),( 1 ,y+z, 0 ),(z, z^2, 1 )|`

Applying `R_1 to R_1-R_2`

` " " " " =(1+pxyz)(x-y)(y-z)|(1-1,x+y-y-1,0 -0),( 1 ,y+z, 0 ),(z, z^2, 1 )|`

` " " " " =(1+pxyz)(x-y)(y-z)|(0,x+y-y-1,0),( 1 ,y+z, 0 ),(z, z^2, 1 )|`

` " " " " =(1+pxyz)(x-y)(y-z)(0[[1,y+z],[z,z^2]]-0[[0,x-z],[z,z^2]]+1[[0,x-z],[1,y+z]])`

` " " " " =(1+pxyz)(x-y)(y-z)[0-0+1(0-(x-z))]`

` " " " " =(1+pxyz)(x-y)(y-z)(z-x)`

` " " " " =` R.H.S.