Prove that `[[x, x^2 , 1+px^3], [y, y^2, 1+py^3] ,[z, z^2, 1+pz^3]] = (1+pxyz)(x-y)(y-z)(z-x)`

Explanation

`|{:(x,x^(2),1+px^(3)),(y,y^(2),1+py^(3)),(z,z^(2),1+pz^(3)):}|=-|{:(x,x^(2),1),(y,y^(2),1),(z,z^(2),1):}|+|{:(x,x^(2),1+px^(3)),(y,y^(2),1+py^(3)),(z,z^(2),1+pz^(3)):}|`
`(C_(1) hArrC_(2)" in first determinant")`
`|{:(x^(2),x,1),(y^(2),y,1),(z^(2),z,1):}|+pxyz|{:(1,x,x^(2)),(1,y,y^(2)),(1,z,z^(3)):}|`
`(C_(1)hArrC_(3)" in first determinant")`
`=(1+pxyz)|{:(1,x,x^(2)),(1,y,y^(2)),(1,z,z^(2)):}|`
`(1+pxyz)|{:(1,x,x^(2)),(0,y-x,y^(2)-x^(2)),(0,z-x,z^(2)-x^(2)):}|`
`(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))`
`=(1+pxyz)(y-x)(z-x)|{:(1,x,x^(2)),(0,1,y+x),(0,1,z+x):}|`
`=-(1+pxyz)(x-y)(z-x)1.|{:(1,y+z),(1,z+x):}|`
=-(1+pxyz)(x-y)(z-x)(z+x-y-x)
=(1+pxyz)(x-y)(y-z)(z-x)=R.H.S.