Show that `{:|(x+2a,y+2b,z+2c),(x,y,z),(a,b,c)|:}`=0

Prove that |(x+2a,y+2b,z+2c),(x,y,z),(a,b,c)|=0

Show that |(1,1,1),(x,y,z),(x^(2),y^(2),z^(2))|=(x-y)(y-z)(z-x)

Show that |(x,y,z),(x^(2),y^(2),z^(2)),(x^(3),y^(3),z^(3))|=xyz (x-y) (y-z) (z-x)

Show that |(x+y,y+z,z+x),(y+z,z+x,x+y),(z+x,x+y,y+z)|=2[3xyz-x^(3)-y^(3)-z^(3)]

If u=ax+by+cz , v=ay+bz+cx , w=ax+bx+cy , then the value of |{:(a,b,c),(b,c,a),(c,a,b):}|xx|{:(x,y,z),(y,z,x),(z,x,y):}| is

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)

Prove that |{:(x^(2),,x^(2)-(y-z)^(2),,yz),(y^(2),,y^(2)-(z-x)^(2),,zx),(z^(2),,z^(2)-(x-y)^(2),,xy):}| =(x-y) (y-z) (z-x)(x+y+z) (x^(2)+y^(2)+z^(2))

Solve that |(y+z,x,x^(2)),(z+x,y,y^(2)),(x+y,z,z^(2))|=(x+y+z)(x-y)(y-z)(z-x)

Using the property of determinants and without expanding {:[( x,a,x+a),( y,b,y+b),(z,c,z+c)]:} =0

If the system of equation {:(,x-2y+z=a),(2x+y-2z=b),and,(x+3y-3z=c):} have at least one solution, then