`|(x,x^(2),1+x^(3)),(y,y^(2),1+y^(3)),(z,z^(2),1+z^(2))|=0,x!=y!=zimplies1+xyz=`

If xneynez" and " |{:(x,x^(2),1+x^(3)),(y,y^(2),1+y^(3)),(z,z^(2),1+z^(3)):}|=0, then xyz =

If xneynez" and " |{:(x,x^(2),1+x^(3)),(y,y^(2),1+y^(3)),(z,z^(2),1+z^(3)):}|=0, then xyz =

Given that xyz = -1 , the value of the determinant |(x,x^(2),1 +x^(3)),(y,y^(2),1 + y^(3)),(z,z^(2),1 +z^(3))| is

If x, y, z are all distinct and |(x,x^(2),1+x^(3)),(y,y^(2),1+y^(3)),(z,z^(2),1+z^(3))|=0 then value of x y z is :

If x,y,z are different and Delta=|{:(x,x^(2),1+x^(3)),(y,y^(2),1+y^(3)),(z,z^(2),1+z^(3)):}|=0 , show that : (i) 1+xyz=0 (ii) xyz=-1

Prove the following : |{:(x,y,z),(x^(2),y^(2),z^(2)),(x^(3),y^(3),z^(3)):}|=|{:(x,x^(2),x^(3)),(y,y^(2),y^(3)),(z,z^(2),z^(3)):}|=xyz(x-y)(y-z)(z-x)

If x,y,z are different and Delta= {:|(x,x^2,1+x^3),(y,y^2,1+y^3),(z,z^2,1+z^3)|=0 , show that xyz=-1

If x,y,z are different and Delta=det[[x,x^(2),1+x^(3)y,y^(2),1+y^(3)z,z^(2),1+z^(3)]]=0, then prove that 1+xyz=0,

If x,y,z are all distinct and if abs((x,x^2,1+x^3),(y,y^2,1+y^3),(z,z^2,1+z^3)) =0,show that xyz+1=0