Question

If `(x + iy)^(1/3)` = a + ib, where x, y, a, b ∈ R, show that `x/a - y/b` = –2(a² + b²)

Answer 1

`(x + iy)^(1/3)` = a + ib

⇒ x + iy = (a + ib)³

i.e., x + iy = a³ + i³ b³ + 3iab (a + ib)

= a³ – ib³ + i3a²b – 3ab²

= a³ – 3ab² + i(3a²b – b³)

⇒ x = a³ – 3ab² and y = 3a²b – b³

Thus `x/a = a^2 - 3b^2` and `y/b = 3a^2 - b^2`

So, `x/a - y/b = a^2 - 3b^2 + b^2`

= `-2a^2 - 2b^2`

= –2(a² + b²)