prove that determinant of x x^{2 } yz

y y^{2} zx = (x-y)(y-z)(z-x)(xy+yz+zx)

z z^{2} xy

First, subtract the second row from the first row. We know thatthe determinant will not change . So, we can say that the aboveexpression (required determinant) equals,As you see, from the first row, we can factor (x-y):Again we can subtract the third row from the second row:Again we can factor (y-z) from the second row:Now, we can subtract the second row from the first. We'll have:We can now factor (x-z) from the first row:We can now subtract z times the first column from the second column:We have to make as many zeros as possible, since they make some parts of the expansion vanish.Now,expand the determinant.= 0 ( ... ) - 1 (xy) + z(-x-y) = -(xy+xz+yz)So all in all, the determinant will be:(x-y)(y-z)(x-z)(-(xy+xz+yz))Let's apply the negative sign in (x-z) and write it as (z-x). We'll have(x - y)(y - z)(z - x)(xy + xz + yz),

**
**