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


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