(x + iy)^{3} = u + iv

x^{3} + (iy)^{3} + 3× x ×iy(x + iy) = u + iv

x^{3} + i^{3}y^{3} + 3x^{2}yi + 3xy^{2}i^{2} = u + iv

x^{3} - iy^{3} + 3x^{2}yi - 3xy^{2} = u + iv

(x^{3} - 3xy^{2}) + i(3x^{2}y - y^{3}) = u + iv

On equating real and imaginary

u = x^{3} - 3xy^{2} ,v = 3x^{2}y - y^{3}

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

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

x^{2} - 3y^{2} + 3x^{2} - y

4x^{2} - 4y^{2}

4(x^{2 }- y^{2})

Hence, proved

Answered by Pragya Singh | 1 year agoShow that 1 + i^{10} + i^{20} + i^{30} is a real number?

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