Prove that $\begin{vmatrix}1&1&1\x&y&z\x^3&y^3&z^3\end{vmatrix}=(y-z)(z-x)(x-y)(x+y+z)$

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> [!TLDR]
> Use the factor theorem: the determinant vanishes whenever any two of $x,y,z$ are equal, so $(x-y)(y-z)(z-x)$ must divide it; matching degrees and one numeric check pins down the remaining factor as $(x+y+z)$.
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> Let $D=\begin{vmatrix}1&1&1\x&y&z\x^3&y^3&z^3\end{vmatrix}$.
>
> **Factor argument:** If $x=y$, columns 1 and 2 become identical ($1,x,x^3$ vs $1,y,y^3$ with $x=y$), so $D=0$. Hence $(x-y)$ is a factor of $D$. By the same reasoning (symmetry of the determinant under swapping any two of $x,y,z$, up to sign), $(y-z)$ and $(z-x)$ are also factors.
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> So $D=(x-y)(y-z)(z-x)\cdot Q(x,y,z)$ for some polynomial $Q$.
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> **Degree count:** row 1 contributes degree 0, row 2 degree 1, row 3 degree 3 to each term of the determinant expansion, so $D$ is homogeneous of degree $0+1+3=4$. Since $(x-y)(y-z)(z-x)$ has degree 3, $Q$ must be degree 1 and — being unchanged in form under any permutation of $x,y,z$ up to the sign already carried by the antisymmetric factor — $Q$ must be a **symmetric** linear expression, i.e. $Q=k(x+y+z)$ for a constant $k$.
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> **Find $k$** using specific values $x=0,y=1,z=2$:
> …

Q.Prove that $\begin{vmatrix}1&1&1\\x&y&z\\x^3&y^3&z^3\end{vmatrix}=(y-z)(z-x)(x-y)(x+y+z)$

Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

The geometric intuition

The three row (or column) operations

The strategy

Worked example

Why determinants change under row operations

Step-by-step evaluation

Q.Prove that ∣111xyzx3y3z3∣=(y−z)(z−x)(x−y)(x+y+z)\begin{vmatrix}1&1&1\\x&y&z\\x^3&y^3&z^3\end{vmatrix}=(y-z)(z-x)(x-y)(x+y+z)​1xx3​1yy3​1zz3​​=(y−z)(z−x)(x−y)(x+y+z)

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Q.Prove that $\begin{vmatrix}1&1&1\\x&y&z\\x^3&y^3&z^3\end{vmatrix}=(y-z)(z-x)(x-y)(x+y+z)$

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