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:
\[ z = \frac{-1 + i\sqrt{3}}{2} \]

### Step 2: Compute \(z^3\)
We will compute \(z^3\) using the formula for the cube of a binomial:
\[ (a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2 \]
Here, let \(a = -\frac{1}{2}\) and \(b = \frac{i\sqrt{3}}{2}\).

### Step 3: Calculate \(a^3\) and \(b^3\)
First, calculate \(a^3\):
\[ a^3 = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8} \]

Next, calculate \(b^3\):
\[ b^3 = \left(\frac{i\sqrt{3}}{2}\right)^3 = \frac{i^3 \cdot (\sqrt{3})^3}{2^3} = \frac{-i \cdot 3\sqrt{3}}{8} = -\frac{3i\sqrt{3}}{8} \]

### Step 4: Calculate \(3a^2b\) and \(3ab^2\)
Now calculate \(3a^2b\):
\[ 3a^2b = 3\left(-\frac{1}{2}\right)^2\left(\frac{i\sqrt{3}}{2}\right) = 3\left(\frac{1}{4}\right)\left(\frac{i\sqrt{3}}{2}\right) = \frac{3i\sqrt{3}}{8} \]

Next, calculate \(3ab^2\):
\[ 3ab^2 = 3\left(-\frac{1}{2}\right)\left(\frac{i\sqrt{3}}{2}\right)^2 = 3\left(-\frac{1}{2}\right)\left(\frac{-3}{4}\right) = \frac{9}{8} \]

### Step 5: Combine all parts
Now combine all the parts:
\[ z^3 = a^3 + b^3 + 3a^2b + 3ab^2 = -\frac{1}{8} - \frac{3i\sqrt{3}}{8} + \frac{3i\sqrt{3}}{8} + \frac{9}{8} \]

The \(3i\sqrt{3}\) terms cancel out:
\[ z^3 = -\frac{1}{8} + \frac{9}{8} = \frac{8}{8} = 1 \]

### Conclusion
Thus, we have shown:
\[ \left(\frac{-1 + i\sqrt{3}}{2}\right)^3 = 1 \]
Since 1 is a positive integer, we conclude that the expression is indeed a positive integer.