using the properties of the determinants prove that det(1 1 1+x1 1+y 11+z 1 1)= xyz+ xy +yz +xz

Det(1 1 1+x 1 1+y 1 1+z 1 1)= by applying row operations that is r1-r2, r2-r3 the matrix becomes (0 -y x -z y 0 1+z 1 1) expanding along c3det =x(y+yz +z ) -0(-y(1+z))+1(yz) =xy+xyx+xz+zy hence proved