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

It is given in the question that,

(x + iy)3 = u + iv

we know that, (x + y)3 = x3 + y3 + 3xy(x + y)

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

x3 + i3y3 + 3x2yi + 3xi2y2= u + iv

Now we know that, i3 = - i and i2 = -1

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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