`{:|( 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 .

For any scalar c prove that, |{:(x,x^2,1+cx^3),(y,y^2,1+cy^3),(z,z^2,1+cz^3):}|=(1+cxyz)(x-y)(y-z)(z-x)

Using properties of determinants porve 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 are different and Delta = {:|( x,x^(2) , 1+x^(3)),( y,y^(3) ,1+y^(3)),( z,z^(3) ,1+z^(3)) |:} then value of delta?

If |[x^n, x^(n+2), x^(n+3)], [y^n, y^(n+2), y^(n+3)], [z^n, z^(n+2), z^(n+3)]|=(x-y)(y-z)(z-x)(1/x+1/y+1/z), then n equals a. 1 b. -1 c. 2 d. -2

Verify that x ^(3) + y ^(3) + z ^(3) - 3xyz =1/2 (x + y + z) [(x-y)^(2) + (y-z) ^(2) + (z-x) ^(2) ]

If |{:(x,x^2,x^3+1),(y,y^2,y^3+1),(z,z^2,z^3+1):}|=a and x ney nez , prove that, xyz=-1.

If x!=y!=za n d|[x,x^2, 1+x^3],[y ,y^2 ,1+y^3],[z, z^2, 1+z^3]|=0, then the value of x y z is a. 1 b. 2 c. -1 d. 2

Factorise : x(1+y^(2))(1+z^(2))+y(1+z^(2))(1+x^(2))+z(1+x^(2))(1+y^(2))+4xyz

If (x_(1),y_(1),z_(1)) , (x_(2),y_(2),z_(2)) , (x_(3) ,y_(3),z_(3)) and (x_(4) , y_(4) , z_(4)) be the consecutive vertices of a parallelogram, show that x_(1)+x_(3)=x_(2)+x_(4),y_(1)+y_(3)=y_(2)+y_(4) and z_(1)+z_(3)=z_(2)+z_(4) .