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)) ]:}` find `|Delta|`.

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 then show that 1 + xyz = 0

using properties of determinant prove that {:[( 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 .

Let Delta=|(Ax,x^(2),1),(By,y^(2),1),(Cz,z^(2),1)| and Delta_(1)=|(A,B,C),(x,y,z),(zy,zx,xy)| then

|[1, x, x^2 ],[1, y, y^2],[1, z, z^2]|=

Delta= |[Ax,x^(2),1],[By,y^(2),1],[Cz, z^(2),1]| and Delta_(1)= |[A,B,C],[x,y,z],[zy, zx,xy]|

If x, y, z are all different and not equal to zero and |{:(1+x,,1,,1),(1,,1+y,,1),(1,,1,,1+z):}| = 0 then the value of x^(-1) + y^(-1) + z^(-1) is equal to

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

Without expanding prove that Delta ={:|( x+y,y+z,z+x) ,( z,x,y),( 1,1,1) |:} =0

Prove that |{:(,x+y+2z,x,y),(,z,y+z+2z,y),(,z,x,z+x+2y):}|=2(x+y+z)^(3) .

The value of the determinant |{:(1 , x , x+ z) , (1 , y , z + x) , (1 , z , x + y):}| is