To Prove:

(x + 1 + i) (x + 1 – i) (x – 1 + i) (x – 1 – i) = (x⁴ + 4)

Taking LHS

(x + 1 + i) (x + 1 – i) (x – 1 + i) (x – 1 – i)

= [(x + 1) + i][(x + 1) – i][(x – 1) + i][(x – 1) – i]

Using (a – b)(a + b) = a² – b²

= [(x + 1)² – (i)²] [(x – 1)² – (i)²]

= [x² + 1 + 2x – i²](x² + 1 – 2x – i²]

= [x² + 1 + 2x – (-1)](x² + 1 – 2x – (-1)] [∵ i² = -1]

= [x² + 2 + 2x][x² + 2 – 2x]

Again, using (a – b)(a + b) = a² – b²

Now, a = x² + 2 and b = 2x

= [(x² + 2)² – (2x)²]

= [x⁴ + 4 + 2(x²)(2) – 4x²] [∵(a + b)² = a² + b² + 2ab]

= [x⁴ + 4 + 4x² – 4x²]

= x⁴ + 4

= RHS

∴ LHS = RHS

Hence Proved