`1/(216)^(-2/3) + 1/(256)^(-3/4) + 1/(243)^(-1/5)` is equal to:

Options

A. 103
B. 105
C. 107
D. 101

Answer

To solve the expression \( \frac{1}{216^{-\frac{2}{3}}} + \frac{1}{256^{-\frac{3}{4}}} + \frac{1}{243^{-\frac{1}{5}}} \), we will simplify each term step by step.

### Step 1: Rewrite the bases as powers
We can express the numbers in the denominators as powers of smaller integers:
- \( 216 = 6^3 \)
- \( 256 = 4^4 \)
- \( 243 = 3^5 \)

So, we rewrite the expression:
\[ \frac{1}{(6^3)^{-\frac{2}{3}}} + \frac{1}{(4^4)^{-\frac{3}{4}}} + \frac{1}{(3^5)^{-\frac{1}{5}}} \]

### Step 2: Apply the power of a power rule
Using the rule \( (a^m)^n = a^{m \cdot n} \):
- For \( 216 \): \( (6^3)^{-\frac{2}{3}} = 6^{3 \cdot -\frac{2}{3}} = 6^{-2} \)
- For \( 256 \): \( (4^4)^{-\frac{3}{4}} = 4^{4 \cdot -\frac{3}{4}} = 4^{-3} \)
- For \( 243 \): \( (3^5)^{-\frac{1}{5}} = 3^{5 \cdot -\frac{1}{5}} = 3^{-1} \)

Now the expression becomes:
\[ \frac{1}{6^{-2}} + \frac{1}{4^{-3}} + \frac{1}{3^{-1}} \]

### Step 3: Simplify the fractions
Using the property \( \frac{1}{a^{-n}} = a^n \):
- \( \frac{1}{6^{-2}} = 6^2 \)
- \( \frac{1}{4^{-3}} = 4^3 \)
- \( \frac{1}{3^{-1}} = 3^1 \)

So now we have:
\[ 6^2 + 4^3 + 3^1 \]

### Step 4: Calculate each term
Now we calculate each power:
- \( 6^2 = 36 \)
- \( 4^3 = 64 \)
- \( 3^1 = 3 \)

### Step 5: Add the results
Now we add these values together:
\[ 36 + 64 + 3 = 103 \]

Thus, the final answer is:
\[ \boxed{103} \]