# prove that determinant of x x2 yz y y2 zx = (x-y)(y-z)(z-x)(xy+yz+zx) z z2 xy

`We want to evaluate:First, subtract the second row from the first row.  We know that the determinant will not change .  So, we can say that the above expression (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),`

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x    x2    yz

y     y2    zx

z     z2    xy

=(x-y)(y-z)(z-x)(xy+yz+zx)

pls tell me the answer fast ty

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Its a determinant and i have to prove it and y2 = y2 and z2=z

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It is a determinant and y2=y2 and z2=z2

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What are you looking for?