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Question
If x + iy = (a + ib)3, show that `x/"a" + y/"b"` = 4(a2 − b2)
Solution
x + iy = (a + ib)3
∴ x + iy = a3 + 3a2(ib) + 3a(ib)2 + (ib)3
∴ x + iy = a3 + 3a2bi + 3ab2i2 + b3i3
∴ x + iy = a3 + 3a2bi – 3ab2 – b3i ...[∵ i2 = – 1, i3 = – i]
∴ x + yi = (a3 – 3ab2) + (3a2b – b3)i
Equating the real and imaginary parts separately, we get,
x = a3 – 3ab2 and y = 3a2b – b3
∴ x = a(a2 – 3b2) and y = b(3a2 – b2)
∴ `x/"a"` = a2 – 3b2 and `y/"b"` = 3a2 – b2
∴ `x/"a" + y/"b"` = a2 – 3b2 + 3a2 – b2 = 4a2 – 4b2
∴ `x/"a" + y/"b"` = 4(a2 – b2)
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