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))`

Explanation

we have
`Delta =|{:(x^(2),,x^(2)-(y-z)^(2),,yz),(y^(2),,y^(2)-(z-x)^(2),,zx),(z^(2),,z^(2)-(x-y)^(2),,xy):}|`
Applying `C_(2) to C_(2) -2 C_(1) -2C_(3)` we get
`Delta =|{:(x^(2),,-(x^(2)+y^(2)+z^(2)),,yz),(y^(2),,-(x^(2)+y^(2)+z^(2)),,zx),(z^(2),,-(x^(2)+y^(2)+z^(2)),,xy):}|`
`=- (x^(2) +Y^(2) +z^(2)) |{:(x^(2),,1,,yz),(y^(2),,1,,zx),(z^(2),,1,,xy):}|`
Multiplying `R_(1),R_(2) " and " R_(3) by x, y` and z, respectively we get
`Delta =- ((x^(2)+y^(2)+z^(2)))/(xyz) |{:(x^(3),,x,,yz),(y^(3),,y,,zx),(z^(3),,z,,xy):}|`
`=- (x^(2) +y^(2) +z^(2)) |{:(x^(3),,x,,1),(y^(3),,y,,1),(z^(3),,z,,1):}| "( Taking xyz common from " c_(3)")"`
`=(x^(2)+y^(2)+z^(2)) |{:(1,,x,,x^(3)),(1,,y,,y^(3)),(1,,z,,z^(3)):}| "(Applying " C_(1) hArr C_(3)")"`
` =(x-y) (y-z) (z-x) (x+y+z) (x^(2) +y^(2)+z^(2))`