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Question
If `(x + iy)^(1/3)` = a + ib, where x, y, a, b ∈ R, show that `x/a - y/b` = –2(a2 + b2)
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Solution
`(x + iy)^(1/3)` = a + ib
⇒ x + iy = (a + ib)3
i.e., x + iy = a3 + i3 b3 + 3iab (a + ib)
= a3 – ib3 + i3a2b – 3ab2
= a3 – 3ab2 + i(3a2b – b3)
⇒ x = a3 – 3ab2 and y = 3a2b – b3
Thus `x/a = a^2 - 3b^2` and `y/b = 3a^2 - b^2`
So, `x/a - y/b = a^2 - 3b^2 + b^2`
= `-2a^2 - 2b^2`
= –2(a2 + b2)
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