Evaluate: `[i^(18) + ((1)/(i))^(25)]^(3)`

Answer

To evaluate the expression \([i^{18} + \left(\frac{1}{i}\right)^{25}]^{3}\), we can follow these steps:

### Step 1: Simplify \(i^{18}\)
We know that the powers of \(i\) cycle every 4:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

To find \(i^{18}\), we can calculate \(18 \mod 4\):
\[ 18 \div 4 = 4 \quad \text{(remainder 2)} \]
Thus, \(i^{18} = i^2 = -1\).

### Step 2: Simplify \(\left(\frac{1}{i}\right)^{25}\)
We can rewrite \(\frac{1}{i}\) as \(-i\) (since \(\frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i\)). Therefore, we need to evaluate \((-i)^{25}\).

Using the same cycle for powers of \(i\):
\[ (-i)^{25} = (-1)^{25} \cdot i^{25} = -i^{25} \]
Now, calculate \(25 \mod 4\):
\[ 25 \div 4 = 6 \quad \text{(remainder 1)} \]
So, \(i^{25} = i^1 = i\), hence:
\[ (-i)^{25} = -i \]

### Step 3: Combine the results
Now we can substitute back into the expression:
\[ i^{18} + \left(\frac{1}{i}\right)^{25} = -1 - i \]

### Step 4: Raise to the power of 3
Now we need to compute:
\[ [-1 - i]^3 \]
Using the binomial theorem:
\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
Here, \(a = -1\) and \(b = -i\):
\[ (-1)^3 + 3(-1)^2(-i) + 3(-1)(-i)^2 + (-i)^3 \]
Calculating each term:
- \((-1)^3 = -1\)
- \(3(-1)^2(-i) = 3(-1)(-i) = 3i\)
- \(3(-1)(-i)^2 = 3(-1)(-1) = 3\)
- \((-i)^3 = -i^3 = -(-i) = i\)

Now combine these results:
\[ -1 + 3i + 3 + i = (2 + 4i) \]

### Final Result
Thus, the final answer is:
\[ \boxed{2 + 4i} \]