Question

If \(i = \sqrt{-1}\) then \([i^{18}+(\frac{1}{i})^{25}]^3 =\)

• A. \(4+4i\)
• B. \(2+2i\)
• C. \(4-4i\)
• D. \(2-2i\)

Hint:

Reduce the powers of i using the cycle of four, then cube.

Answer 1

The Correct Option is D

Step 1: Approach
Convert everything to a single power of $i$ first.

Step 2: Combine
$\left(\dfrac1i\right)^{25}=i^{-25}$. Then $i^{-25}=i^{-25+28}=i^{3}=-i$ because $i^{28}=1$.
Also $i^{18}=i^{2}=-1$.

Step 3: Sum
$-1-i$. To cube it, use the binomial form:
\[ (-1-i)^3=-(1+i)^3=-(1+3i+3i^2+i^3)=-(1+3i-3-i) \]
\[ =-(-2+2i)=2-2i \]
This matches option (D).

Final Answer:
The bracket equals $-1-i$, and its cube is $2-2i$, option (D). \[ \boxed{2-2i} \]