Simplify : `((8)/(27))^(-1//3)xx((25)/(4))^(1//2)xx((4)/(9))^(0)+((125)/(64))^(1//3)`

Answer

To simplify the expression \(\left(\frac{8}{27}\right)^{-\frac{1}{3}} \times \left(\frac{25}{4}\right)^{\frac{1}{2}} \times \left(\frac{4}{9}\right)^{0} + \left(\frac{125}{64}\right)^{\frac{1}{3}}\), we will follow these steps:

### Step 1: Simplify each term using properties of exponents
1. **For \(\left(\frac{8}{27}\right)^{-\frac{1}{3}}\)**:
- Rewrite \(8\) and \(27\) as powers: \(8 = 2^3\) and \(27 = 3^3\).
- Thus, \(\frac{8}{27} = \frac{2^3}{3^3} = \left(\frac{2}{3}\right)^3\).
- Therefore, \(\left(\frac{8}{27}\right)^{-\frac{1}{3}} = \left(\left(\frac{2}{3}\right)^3\right)^{-\frac{1}{3}} = \left(\frac{2}{3}\right)^{-1} = \frac{3}{2}\).

2. **For \(\left(\frac{25}{4}\right)^{\frac{1}{2}}\)**:
- Rewrite \(25\) and \(4\) as powers: \(25 = 5^2\) and \(4 = 2^2\).
- Thus, \(\frac{25}{4} = \frac{5^2}{2^2} = \left(\frac{5}{2}\right)^2\).
- Therefore, \(\left(\frac{25}{4}\right)^{\frac{1}{2}} = \left(\left(\frac{5}{2}\right)^2\right)^{\frac{1}{2}} = \frac{5}{2}\).

3. **For \(\left(\frac{4}{9}\right)^{0}\)**:
- Any number raised to the power of \(0\) is \(1\).
- Thus, \(\left(\frac{4}{9}\right)^{0} = 1\).

4. **For \(\left(\frac{125}{64}\right)^{\frac{1}{3}}\)**:
- Rewrite \(125\) and \(64\) as powers: \(125 = 5^3\) and \(64 = 4^3\).
- Thus, \(\frac{125}{64} = \frac{5^3}{4^3} = \left(\frac{5}{4}\right)^3\).
- Therefore, \(\left(\frac{125}{64}\right)^{\frac{1}{3}} = \left(\left(\frac{5}{4}\right)^3\right)^{\frac{1}{3}} = \frac{5}{4}\).

### Step 2: Substitute back into the expression
Now substituting the simplified terms back into the expression:
\[ \frac{3}{2} \times \frac{5}{2} \times 1 + \frac{5}{4} \]

### Step 3: Calculate the first part
Calculate \(\frac{3}{2} \times \frac{5}{2}\):
\[ \frac{3 \times 5}{2 \times 2} = \frac{15}{4} \]

### Step 4: Add the two parts together
Now add \(\frac{15}{4}\) and \(\frac{5}{4}\):
\[ \frac{15}{4} + \frac{5}{4} = \frac{15 + 5}{4} = \frac{20}{4} = 5 \]

### Final Answer
Thus, the simplified expression is:
\[ \boxed{5} \]