`|{:(x+y+z," "-z," "-y),(" "-z,x+y+z," "-x),(" "-y," "-x,x+y+z):}|=2(x+y)(y+z)(z+x)`

|{:(y+z," "z," "y),(" "z,z+x," "x),(" "y," "x,x+y):}|=4xyz

|{:(" "1," "z,-y),(-z," "1," "x),(" "y,-x," "1):}|

|{:(a+x," "y," "z),(" "x,a+y," "z),(" "x," "y,a+z):}|=a^2(a+x+y+z)

|{:(" "0," "x,y),(-x," "0,z),(-y,-z,0):}|=0

By using 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)

Simplify: x(x-y)+y(y-z)+z(z-x)

The three adjacent surfaces of cuboid are x sq. unit, y sq. unit and z sq. unit. The length of its one of the diagonal is Option 1 (xy/z+yz/x+zx/y)^(1/2) Option 2 ( z x/ y + x y/ z + y z/ x )^(1/2) Option 3 ( x y/ z + y z/ x + z x/ y ) Option 4 ( z y/ x + y z/ x + x y/ z )

(x-y)^(3)+(y-z)^(3)+(z-x)^(3)

Prove that |[a x-b y-c z, a y+b x, c x+a z], [a y+b x, b y-c z-a x, b z+c y],[c x+a z, b z+c y, c z-a x-b y]|=(x^2+y^2+z^2)(a^2+b^2+c^2)(a x+b y+c z)dot

If x,y,z in Z then (x +y) z = x, z +y , z - this law is called