Using properties of determinants, prove the following:

Taking LHS,

By applying R₁→ R₁ + R₂ + R₃, we get

Taking (1 + x + x²) common from the first row, we get

Applying R₁→ R₁ – R₂, we get

Applying R₂→ R₂ – R₃, we get

Taking common (1 – x) from second row, we get

Applying R₂→ R₂ + R₁, we get

Applying R₃→ R₃ – R₁, we get

Expanding along C₃, we get

= (1 + x + x²)(1 – x)² [(1){(x²) – (-1)(1 + x)}]

= (1 + x + x²)(1 – x)² [x² + 1 + x]

= (1 + x + x²)(1 – x)²(x² + x + 1)

= (1 – x)(1 + x + x²)(1 – x)(1 + x + x²)

= {1 + x + x² – x – x² – x³)²

= (1 – x³)²

= RHS

Hence,

∴ LHS = RHS

Hence Proved