Discussion
If \(i = \sqrt{-1}\), \(\dfrac{(i+1)^{3}}{i-1}\) = ________.
Answer: A
The value is 2.
Expand the cube using \(i^{2} = -1\) and \(i^{3} = -i\), then divide:
\[ \begin{aligned} (i + 1)^{3} &= i^{3} + 3i^{2} + 3i + 1 \\ &= -i - 3 + 3i + 1 = 2i - 2 \\ \frac{2i - 2}{i - 1} &= \frac{2(i - 1)}{i - 1} = 2 \end{aligned} \]
A quicker route: \((1 + i)^{2} = 2i\), so \((1 + i)^{3} = 2i(1 + i) = 2i - 2\). The answer is the real number 2. −2 comes from a sign slip in \(i^{3}\) or in the denominator, and \((i + 1)\) and \((i - 1)\) leave the result complex, which it is not.
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