|[px+y,x,x],[py+z,y,z],[0,px+y,py+z]|=0

If x,y,z are in GP, then using properties,det[[px+y,x,ypy+z,y,z0,px+y,py+z]]=0 where x!=y!=z and p is any real number.

If Delta= [[0 , x , -y ],[-x , 0 , z],[ y , -z , 0 ]] then

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

Using properties of determinants, prove the following: |[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)

Using properties of determinants, prove the following: |[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)

Without expanding show that : |(x-y,y-z,z-x),(y-z,z-x,x-y),(z-x,x-y,y-z)|=0

|{:(x,z,0),(0,y,y),(z,0,x):}|

Using properties of determinants, 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 = 0 then prove that |{:(x,y,z),(x^2,y^2,z^2),(y+z,z+x,x+y):}|=0

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)