For any scalar `p` prove that `=|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)` .

For any scalar p 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) .

For any scalar p prove that =|x^(2)1+px^(3)yy^(2)1+py^(3)zz^(2)1+pz^(3)|=(1+pxyz)(x-y)(y-z)(z-x)

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)

If x!=y!=z and |x x^2 1+x^3 y y^2 1+y^3 z z^2 1+z^3|=0 , then prove that x y z=-1 .

The determinant |x p+y x y y p+z y z0x p+y y p+z|=0 if x ,y ,z

The determinant |x p+y x y y p+z y z0x p+y y p+z|=0 if x ,y ,z

{:[( 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) , where p is any scalar .

{:|( x,x^(2) , 1+ px^(3) ),( y,y^(2) , 1+ py^(2)),( z,z^(2) , 1+pz^(2)) |:} =( 1+pxyz ) ( x-y) ( y-z ) (z-x) , where p is any scalar .