`((13)^(3)+7^(3))/((13)^(2)+7^(2)-?)=20`

Options

A. 6
B. 20
C. 91
D. 104

Explanation

To solve the equation \(\frac{13^3 + 7^3}{13^2 + 7^2 - ?} = 20\), we will follow these steps:

### Step 1: Write the equation
We start with the equation:
\[ \frac{13^3 + 7^3}{13^2 + 7^2 - x} = 20 \]

### Step 2: Apply the formula for \(A^3 + B^3\)
Recall the formula for the sum of cubes:
\[ A^3 + B^3 = (A + B)(A^2 - AB + B^2) \]
Here, let \(A = 13\) and \(B = 7\). Thus:
\[ 13^3 + 7^3 = (13 + 7)(13^2 - 13 \cdot 7 + 7^2) \]
Calculating \(13 + 7\):
\[ 13 + 7 = 20 \]
Now we need to calculate \(13^2\), \(7^2\), and \(13 \cdot 7\):
\[ 13^2 = 169, \quad 7^2 = 49, \quad 13 \cdot 7 = 91 \]
Now substitute these values into the formula:
\[ 13^3 + 7^3 = 20(169 - 91 + 49) = 20(127) \]
Thus:
\[ 13^3 + 7^3 = 2540 \]

### Step 3: Substitute back into the equation
Now we substitute \(13^3 + 7^3\) back into the equation:
\[ \frac{2540}{13^2 + 7^2 - x} = 20 \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 2540 = 20(13^2 + 7^2 - x) \]

### Step 5: Divide both sides by 20
Now divide both sides by 20:
\[ \frac{2540}{20} = 13^2 + 7^2 - x \]
Calculating the left side:
\[ 127 = 13^2 + 7^2 - x \]

### Step 6: Substitute \(13^2 + 7^2\)
We already calculated \(13^2 + 7^2\):
\[ 13^2 + 7^2 = 169 + 49 = 218 \]
Substituting this into the equation:
\[ 127 = 218 - x \]

### Step 7: Solve for \(x\)
Rearranging gives:
\[ x = 218 - 127 \]
Calculating the right side:
\[ x = 91 \]

### Final Answer
The value of \(x\) is:
\[ \boxed{91} \]