Question

Solve it If ( x + iy)⅓ = a + ib ( x , y , a , b £ R )Show that x/a + y/b = 4 (a² - b²)​

Answer 1

$<b> Heya! </b>$

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$<u> ★Complex Number Systems★ </u>$

Given that ,

$= > (x + iy) {}^{ \frac{1}{3} } = a \: + ib \\ \\ = > (x + iy) = (a + ib) {}^{3}$

=> x + iy = a³ + 3a²ib + 3a i²b² + i³b³

=> x + iy = ( a³ - 3ab²) + i (3a²b - b³)

•°• x/a = a² - 3b² and y/b = 3a² - b²

$<b> <u> Adding the two values </u> </b>$

→› x/a + y/b = a² - 3b² + 3a² - b²

→› x/a + y/b = 4a² - 4b²

=> 4 ( a² - b²)

$<b> <u> L.H.S = R.H.S </b> </u>$

Hence Proved!!

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Answer 2

Hello!

( x + iy)^1/3 = (a + ib)

So cubing both the sides

( x + iy) = (a+ib)³

( x + iy) = a³ + i³b³ + 3a²ib + 3ai²b²

Then You can separate the real and imaginary parts

x = a³ - 3ab²

y = b³ + 3a²b

Put the values x/a + y/b

= 4a² - 4b²

= 4 (a² - b²)

Hope it the the correct answer friend.☺

$Hope it is helpful$