# Finding a simple expression for this determinant

Evaluate $$\Delta=\left\lvert\matrix{ 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy }\right\rvert$$

The answer of the above question is $(x-y)(y-z)(z-x)$.

But will solving I get stuck to $-yx^2+zx^2+xy^2-xz^2-zy^2-yz^2$.

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Hint: Remember that $x^2 - y^2 = (x+y)(x-y)$. – MJD Sep 18 '12 at 15:26
Also, you have the wrong sign on $yz^2$. It is supposed to be plus, and you have minus. An easy way to notice that something is wrong is to see that you have two plus terms and four minus terms, but it should be three of each. – MJD Sep 18 '12 at 15:27

Whenever two of the variables are equal, two of the rows are equal, so the determinant is zero. Thus the determinant must contain each of the three differences of the variables as linear factors. Since the determinant is a cubic polynomial, it must be of the form $k(x-y)(y-z)(z-x)$ with $k\in\mathbb R$, and comparing coefficients e.g. of $xy^2$ yields $k=1$.

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$$\Delta=\left\lvert\matrix{ 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy }\right\rvert$$

Transforming $R_1\to R_1-R_2$ and $R_3\to R_3-R_2$gives

$$\Delta =\left\lvert\matrix{ 0 & x-y & -(x-y)z \\ 1 & y & zx \\ 0 & z-y & -x(y-z) }\right\rvert= (x-y)(z-y)\left\lvert\matrix{ 0 & 1 & -z \\ 1 & y & zx \\ 0 & 1 & -x }\right\rvert$$ (Taking $(x-y)$ common out of $R_1$ and $(z-y)$ common out of $R_3$)

Now expand about first column gives $\Delta=(x-y)(z-y)(-z-(-x))=(x-y)(z-y)(x-z)=(x-y)(y-z)(z-x)$

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