solve `|[1,yz,yz(y+z)],[1,zx,zx(z+x)],[1,xy,xy(x+y)]|`

Using Cofactors of elements of third column, evaluate Delta=|[1 , x, yz],[1, y, zx],[1, z, xy]|

show tha |[1,x,x^2-yz],[1,y,y^2-zx],[1,z,z^2-xy]| =0

Prove that : |{:(1,x,yz),(1,y,zx),(1,z,xy):}|=(x-y)(y-z)(z-x)

proof |[x,y,z],[x^(2),y^(2),z^(2)],[yz,zx,xy]| = |[1,1,1],[x^(2),y^(2),z^(2)],[x^(3),y^(3),z^(3)]|

prove that: |(y^(2)z^(2),yz,y+z),(z^(2)x^(2),zx,z+x),(x^(2)y^(2),xy,x+y)|=0

Using the properties of determinants, show that: [[x, x^2, yz],[y, y^2, zx],[z, z^2, xy]]=(x-y)(y-z)(z-x)(xy+yz+zx)

Show that Delta=|((y+z)^2,xy,zx),(xy,(x+z)^2,yz),(xz,yz,(x+y)^2)|=2x y z(x+y+z)^3 .

Prove that |[yz-x^2,zx-y^2,xy-z^2],[zx-y^2,xy-z^2,yz-x^2],[xy-z^2,yz-x^2,zx-y^2]| is divisible by (x+y+z), and hence find the quotient.

If D_1=|[1, 1, 1],[x^2,y^2,z^2],[x, y, z]| and D_2=|[1, 1, 1],[yz, zx,xy], [x, y, z]| without expanding prove that D_1=D_2

If "Delta"=|(1,x,x^2),( 1,y, y^2),( 1,z, z^2)| , "Delta"_1=|(1, 1, 1),(yz, z x,x y), (x, y, z)| , then prove that "Delta"+"Delta"_1=0 .