If x + y + z = 0 then show that |1 1 1; x y z; x³ y³ z³| = 0, using properties of determinant.

If $x + y + z = 0$ then show that $\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^3 & y^3 & z^3 \end{vmatrix} = 0$, using properties of determinant.

Given: x + y + z = 0

To show: $\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^3 & y^3 & z^3 \end{vmatrix} = 0$

Answer

Proof that the determinant is $0$ when $x+y+z=0$

1. Let $\Delta=\begin{vmatrix}1&1&1\\x&y&z\\x^3&y^3&z^3\end{vmatrix}$.
2. Apply $C_2\to C_2-C_1$, $C_3\to C_3-C_1$: $\Delta=\begin{vmatrix}1&0&0\\x&y-x&z-x\\x^3&y^3-x^3&z^3-x^3\end{vmatrix}$.
3. Expand along $R_1$: $\Delta=(y-x)(z^3-x^3)-(z-x)(y^3-x^3)$.
4. Take out $(y-x)(z-x)$: $\Delta=(y-x)(z-x)\left[(z^2+zx+x^2)-(y^2+xy+x^2)\right]$.
5. $(z^2-y^2)+x(z-y)=(z-y)(z+y+x)$, so $\Delta=(y-x)(z-x)(z-y)(x+y+z)$.
6. Since $x+y+z=0$, $\Delta=0$.