Given: (1 + i)3 – (1 – i)3 …(i)
We know that,
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
By applying the formulas in eq. (i), we get
(1)3 + 3(1)2 (i) + 3(1)(i)2 + (i)3 – [(1)3 – 3(1)2 (i) + 3(1)(i)2 – (i)3]
= 1 + 3i + 3i2 + i3 – [1 – 3i + 3i2 – i3]
= 1 + 3i + 3i2 + i3 – 1 + 3i – 3i2 + i3
= 6i + 2i3 = 6i + 2i(i2)
= 6i + 2i(-1) [∵ i2 = -1]
= 6i – 2i = 4i = 0 + 4i
