Using the properties of determinants, prove that:

1 x x³

1 y y³ =(x-y) (y-z) (z-x) (x+y+z)

1 z z³

R21(-1),R32(-1)
so you get 
0  x-y  x^3-y^3
0  y-z  y^3-z^3
0  z      z^3

now using the formula a^3-b^3=(a-b)(a^2+b^2+ab) expand the last terms of 1st and 2nd row

0  x-y  (x-y)(x^2+y^2+xy)
0  y-z  (y-z)(y^2+z^2+yz)
0  z       z^3

now R1(1/x-y),R2(1/y-z)

(x-y)(y-z) | 0    1     x^2+y^2+xy
               | 0     1    y^2+z^2+yz
               | 0     z      z^3
now again R21(-1)
(x-y)(y-z)  0    0     x^2-z^2 +y(x-z)
                0     0    y^2+z^2+yz
                0     z        z^3
take x-z common from first row and then expand