Find the product of square of `(-1)/(2)` and the cube of `(-2)/(3)`.

Answer

To find the product of the square of \(-\frac{1}{2}\) and the cube of \(-\frac{2}{3}\), we can follow these steps:

### Step 1: Calculate the square of \(-\frac{1}{2}\)
The square of a number \(x\) is given by \(x^2\). Therefore, we calculate:
\[ \left(-\frac{1}{2}\right)^2 = \left(-1\right)^2 \cdot \left(\frac{1}{2}\right)^2 = 1 \cdot \frac{1}{4} = \frac{1}{4} \]

### Step 2: Calculate the cube of \(-\frac{2}{3}\)
The cube of a number \(y\) is given by \(y^3\). Therefore, we calculate:
\[ \left(-\frac{2}{3}\right)^3 = \left(-2\right)^3 \cdot \left(\frac{1}{3}\right)^3 = -8 \cdot \frac{1}{27} = -\frac{8}{27} \]

### Step 3: Find the product of the results from Step 1 and Step 2
Now, we need to find the product of \(\frac{1}{4}\) and \(-\frac{8}{27}\):
\[ \frac{1}{4} \cdot \left(-\frac{8}{27}\right) = -\frac{1 \cdot 8}{4 \cdot 27} = -\frac{8}{108} \]

### Step 4: Simplify the fraction
Now, we simplify \(-\frac{8}{108}\):
\[ -\frac{8}{108} = -\frac{2}{27} \quad \text{(dividing both numerator and denominator by 4)} \]

### Final Answer
Thus, the product of the square of \(-\frac{1}{2}\) and the cube of \(-\frac{2}{3}\) is:
\[ -\frac{2}{27} \]