Show that $$\operatorname{det}\;\begin{bmatrix} 1&1&1\\x^2&y^2&z^2\\x^4&y^4&z^4 \end{bmatrix}=(y^2-x^2)(z^2-x^2)(z^2-y^2)$$
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$\operatorname{det}\;\begin{bmatrix} 1&1&1\\x^2&y^2&z^2\\x^4&y^4&z^4 \end{bmatrix} \\=\operatorname{det}\;\begin{bmatrix} 1&0&0\\x^2&y^2-x^2&z^2-x^2\\x^4&y^4-x^4&z^4-x^4 \end{bmatrix}\tag*{}$ (using the transformation $C_2\rightarrow C_2-C_1 $ and $C_3\rightarrow C_3-C_1$) $\\=(y^2-x^2)(z^4-x^4)-(y^4-x^4)(z^2-x^2) \\=(y^2-x^2)(z^2-x^2)((z^2+x^2)-(y^2+x^2)) \\=(y^2-x^2)(z^2-x^2)(z^2-y^2)\tag*{}$
You are given a matrix with entries in some polynomial ring, say $\Bbb Z[x,y,z]$ or $\Bbb R[x,y,z]$ (in this specific case it does not really matter, what the underlying ring is). That being said, you can compute the determinant of a matrix just like any other matrix, for example (since we are given a $3\times 3$ matrix) via Sarrus‘ rule: $$\operatorname{det}\;\begin{bmatrix} 1&1&1\\x^2&y^2&z^2\\x^4&y^4&z^4 \end{bmatrix}=y^2z^4 + z^2x^4 + x^2y^4 - x^4y^2 - y^4z^2 - z^4x^2$$ Now expanding the term of the desired result should give you this as well.