x + iy = (a + ib)^{3}

x + iy = a^{3} + 3a^{2} bi + 3ab^{2} i^{2} + b^{3} i^{3}

x + iy = a^{3} + 3a^{2} bi – 3ab^{2} – b^{3} i ……**[∵ i**^{2} = -1, i^{3} = -i]

x + iy = (a^{3} – 3ab^{2} ) + (3a^{2}b – b^{3} )i

Equating real and imaginary parts, we get

**Alternate Method:**

Equating real and imaginary parts, we get

x = a^{3} – 3ab^{2} and y = 3a^{2} b – b^{3}

**Consider**