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Question
Show that `(-1 + sqrt(3)"i")^3` is a real number.
Solution
`(-1 + sqrt(3)"i")^3`
= `(-1)^3 + 3(-1)^2 (sqrt(3)"i") + 3(-1) (sqrt(3)"i")^2 + (sqrt(3)"i")^3` ...[(a + b)3 = a3 + 3a2b + 3ab2 + b3]
= `- 1 + 3sqrt(3)"i" - 3(3"i"^2) + 3sqrt(3)"i"^3`
= `-1 + 3sqrt(3)"i" - 3(-3) - 3sqrt(3)"i"` ...[∵ i2 = –1, i3 = – i]
= – 1 + 9
= 8, which is a real number.
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