Question

(x+iy)⅓=a+ib then x/a+y/b=​

Answer 1

Answer:

x+iy = a^3- ib^3

x/+y/b= a^2 + ib2

Answer 2

Answer:

${(x + iy)}^{ \frac{1}{3} } = a + ib \\ cubing \: both \: sides \\ x + iy = {(a + ib)}^{3} \\x + iy = {a}^{3} + {(ib)}^{3} + 3 {a}^{2} bi + 3 a {(ib)}^{2} \\ x + iy = {a}^{3} - {ib}^{3} + 3 {a}^{2} bi - 3a {b}^{2} \\ x + iy = {a}^{3} - 3 a{b}^{2} + i( 3 {a}^{2} b - {b}^{3} )$

On comparing real and imaginary values

x = a³ - 3ab²

x = a ( a² - 3b² )

x / a = a² - 3b²......{1}

Also

y = 3a²b - b³

y = b ( 3a²-b²)

y / b = 3a² - b²......{2}

Add {1} & {2}

x/a + y/b = a²- 3b² + 3a² - b²

x/a + y/b = 4a² - 4b²

Here is the answer........

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