Question

If x + y + z = 0 then show that `|(1, 1, 1),(x, y, z),(x^3, y^3, z^3)| = 0`, using properties of determinant.

Answer 1

`Δ = |(1, 1, 1),(x, y, z),(x^3, y^3, z^3)|`

By applying

C₁ → C₁ – C₂

C₂ → C₂ – C₃

⇒ `Δ = |(0, 0, 1),(x - y, y - z, z),(x^3 - y^3, y^3 - z^3, z^3)|`

⇒ `Δ = |(0, 0, 1),(x - y, y - z, z),((x - y), (y - z), z^3),((x^2 + y^2 + xy),(y^2 + z^2 + yz),)|`

⇒ Δ = (x – y) (y – z)

`|(0, 0, 1),(1, 1, z),(x^2 + y^2 + xy, y^2 + z^2 + yz, z^3)|`

⇒ Δ = (x – y) (y – z) [(y² + z² + yz) – (x² + y² + xy)]

⇒ Δ = (x – y) (y – z) [z² – x² + y(z – x)]

⇒ Δ = (x – y) (y – z) [(z – x) (z + x) + y(z – x)]

⇒ Δ = (x – y) (y – z) (z – x) (x + y + z)

But it is given that

x + y + z = 0

∴ Δ = 0.

Hence Proved.