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

(x + iy)³ = u + iv

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

x³ + i³y³ + 3x²yi + 3xy²i² = u + iv

x³ - iy³ + 3x²yi - 3xy² = u + iv

(x³ - 3xy²) + i(3x²y - y³) = u + iv

On equating real and imaginary

u = x³ - 3xy² ,v = 3x²y - y³

\( \dfrac{u}{x}+\dfrac{v}{y}=\dfrac{x^3-3xy^2}{x}+\dfrac{3x^2y-y^3}{y}\)

^{\( \dfrac{x(x^2-3y^2)}{x}+\dfrac{y(3x^2-y^2)}{y}\)}

x² - 3y² + 3x² - y

4x² - 4y²

4(x² - y²)

Hence, proved