Q. 164.4( 54 Votes )

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

Answer :

It is given in the question that,

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

**(x + y)**

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

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

**Now we know that, i**

^{3}= - i and i^{2}= -1x^{3} – iy^{3} + 3x^{2}yi – 3xy^{2} = u + iv

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

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

u = (x^{3} – 3xy^{2}), v = (-y^{3} + 3x^{2}y)

According to question,

= x^{2} – 3y^{2} + 3x^{2} – y^{2}

= 4x^{2} – 4y^{2}

= 4(x^{2} – y^{2})

Thus,

**Hence, proved**

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