If (x + iy)³ – u + iv, then show that:

It is given in the question that,

(x + iy)³ = u + iv

x³ + i³y³ + 3 × x × iy(x + iy) = u + iv

x³ + i³y³ + 3x²yi + 3xi²y²= u + iv

x³ – iy³ + 3x²yi – 3xy² = u + iv

(x³ – 3xy²) + i(-y³ + 3x²y)= u + iv

Now, equating the imaginary and real part we will get,

u = (x³ – 3xy²), v = (-y³ + 3x²y)

According to question,

= x² – 3y² + 3x² – y²

= 4x² – 4y²

= 4(x² – y²)

Thus,

Hence, proved