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

Answer

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

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

To find \(i^{18}\), we can reduce the exponent modulo 4:
\[ 18 \mod 4 = 2 \]
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 calculate:
\[ (-i)^{25} \]
Again, we reduce the exponent modulo 4:
\[ 25 \mod 4 = 1 \]
Thus,
\[ (-i)^{25} = (-i)^1 = -i \]

### Step 3: Combine the results
Now we can combine the results from Step 1 and Step 2:
\[ i^{18} + \left(\frac{1}{i}\right)^{25} = -1 + (-i) = -1 - i \]

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

Combining these results:
\[ (-1 - i)^3 = -1 - 3i + 3 + i = 2 - 2i \]

### Final Answer
Thus, the simplified expression is:
\[ \boxed{2(1 - i)} \]