If z = x – i y and z^{1/3} = p+ iq , then is equal to

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2

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B.
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D.
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therefore,(Qz = x − iy)(x  iy) = (p + iq)^{3}
⇒ (x  iy) = p^{3} +(iq)^{3} + 3p^{2}qi + 3pq^{2}i^{2}
⇒ (x  iy) = p^{3}  iq^{3} + 3p^{2}qi  3pq^{2}
⇒ (x  iy) = (p^{3}  3pq^{2} ) + i (3p^{2} q  q^{3} ) On comparing both sides, we get
⇒ x = (p^{3}  3pq^{2}) and  y = 3p^{2} q  q^{3}
⇒ x = p(p^{2}  3q^{2} ) and y = q(q^{2}  3p^{2} )