If $x \neq y \neq z$ then $\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$ is equal to

The determinant to evaluate is $\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$

This is a Vandermonde Determinant.

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Expanding along the first row:

$\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$

$= 1 \begin{vmatrix} y & y^2 \\ z & z^2 \end{vmatrix} - x \begin{vmatrix} 1 & y^2 \\ 1 & z^2 \end{vmatrix} + x^2 \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix}$

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Evaluating each 2×2 determinant:

$\begin{vmatrix} y & y^2 \\ z & z^2 \end{vmatrix}$

$= yz^2 - y^2z$

$= yz(z - y)$

$\begin{vmatrix} 1 & y^2 \\ 1 & z^2 \end{vmatrix}$

$= z^2 - y^2$

$\begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix}$

$= z - y$

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Substituting back:

$= 1[yz(z-y)] - x[z^2 - y^2] + x^2[z - y]$

$= yz(z-y) - x(z^2 - y^2) + x^2(z - y)$

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Using $z^2 - y^2 = (z-y)(z+y)$:

$= yz(z-y) - x(z-y)(z+y) + x^2(z-y)$

Factoring out $(z - y)$:

$= (z-y)[yz - x(z+y) + x^2]$

$= (z-y)[yz - xz - xy + x^2]$

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Regrouping terms:

$= (z-y)[x^2 - xy - xz + yz]$

$= (z-y)[x(x-y) - z(x-y)]$

$= (z-y)(x-y)(x-z)$

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Rearranging:

$(z-y)(x-y)(x-z)$

$= (x-y)(-(y-z))(-(z-x))$

$= (x-y)(y-z)(z-x)$

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The condition $x \neq y \neq z$ ensures the determinant is non-zero, as equality of any two variables would make one factor zero.

Therefore, $\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = (x-y)(y-z)(z-x)$