Prove that `((-1+isqrt(3))/(2))^(3)` is a positive integer.

Answer

To prove that \(\left(\frac{-1 + i\sqrt{3}}{2}\right)^3\) is a positive integer, we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \left(\frac{-1 + i\sqrt{3}}{2}\right)^3 = \frac{(-1 + i\sqrt{3})^3}{2^3} = \frac{(-1 + i\sqrt{3})^3}{8} \]

### Step 2: Apply the binomial expansion
Using the binomial expansion for \((a + b)^3\), where \(a = -1\) and \(b = i\sqrt{3}\):
\[ (-1 + i\sqrt{3})^3 = (-1)^3 + 3(-1)^2(i\sqrt{3}) + 3(-1)(i\sqrt{3})^2 + (i\sqrt{3})^3 \]
Calculating each term:
- \((-1)^3 = -1\)
- \(3(-1)^2(i\sqrt{3}) = 3 \cdot 1 \cdot i\sqrt{3} = 3i\sqrt{3}\)
- \(3(-1)(i\sqrt{3})^2 = 3(-1)(-3) = 9\) (since \((i\sqrt{3})^2 = -3\))
- \((i\sqrt{3})^3 = i^3 \cdot 3\sqrt{3} = -i \cdot 3\sqrt{3} = -3i\sqrt{3}\)

### Step 3: Combine the terms
Now we combine all these terms:
\[ (-1 + i\sqrt{3})^3 = -1 + 3i\sqrt{3} + 9 - 3i\sqrt{3} \]
The \(3i\sqrt{3}\) and \(-3i\sqrt{3}\) cancel each other out:
\[ (-1 + i\sqrt{3})^3 = 8 \]

### Step 4: Substitute back into the expression
Now substituting back into our original expression:
\[ \frac{(-1 + i\sqrt{3})^3}{8} = \frac{8}{8} = 1 \]

### Conclusion
Since \(1\) is a positive integer, we have proved that:
\[ \left(\frac{-1 + i\sqrt{3}}{2}\right)^3 \text{ is a positive integer.} \]

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