Given that, (x + iy)^1/3 = (a + ib)

⇒ (x + iy) = (a + ib)³

⇒ (a + ib)³ = x + iy

⇒ a³ + (ib)³ + 3a²ib + 3ai²b² = x + iy

⇒ a³ - ib³ + 3a²ib - 3ab² = x + iy

⇒ a³ - 3ab² + i(3a²b - b³) = x + iy

On equating real and imaginary parts, we get

x = a³ - 3ab² and y = 3a²b - b³

Now ,

= a² - 3b² + 3a² - b²

= 4a² - 4b²

= 4(a² - b²)

Hence,