Question

pls solve this pls pls

releted with complex numbers​

Answer 1

Answer:

$(x + iy) {}^{ \frac{1}{3} } = a + ib \\ cubing \: both \: sides \\ x + iy = (a + ib) {}^{3} \\ expanding \: rhs \\ a {}^{3} - b {}^{3} i + 3 {a}^{2} bi - 3a {b}^{2} \\ = (a {}^{3} - 3a {b}^{2} ) + i(3 {a}^{2} b - b {}^{3} ) \\ comparing \: real \: and \: img \: parts \\ x = a {}^{3} - 3a {b}^{2} ......(1)\\ y = 3a {}^{2} b - b {}^{3} .....(2)\\ we \: need \: to \: prove \: \\ \frac{x}{a} + \frac{y}{b} = 4( {a}^{2} - {b}^{2} ) \\ expanding \: lhs \: with \: the \\ \: help \: of \: eqn \: (1)and(2) \\ \frac{ {a}^{3} - 3a {b}^{2} }{a} + \frac{3 {a}^{2} b - b {}^{3} }{b} \\ = {a}^{2} - 3 {b}^{2} + 3 {a}^{2} - {b}^{2} \\ = 4( {a}^{2} - {b}^{2} )$

hence, proved.

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