x and y are positive integers. If ` x^(4) + y^(4) + x^(2) y^(2) = 481 and xy = 12 ` then what is the value of ` x^(2) - xy + y^(2)` ?

Options

A. 16
B. 13
C. 11
D. 15

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Use the given information
We are given two equations:
1. \( x^4 + y^4 + x^2y^2 = 481 \)
2. \( xy = 12 \)

### Step 2: Find \( x^2y^2 \)
From the second equation, we can find \( x^2y^2 \) by squaring both sides of \( xy = 12 \):
\[ (xy)^2 = 12^2 \]
\[ x^2y^2 = 144 \]

### Step 3: Substitute \( x^2y^2 \) into the first equation
Now, substitute \( x^2y^2 = 144 \) into the first equation:
\[ x^4 + y^4 + 144 = 481 \]
Subtract 144 from both sides:
\[ x^4 + y^4 = 481 - 144 \]
\[ x^4 + y^4 = 337 \]

### Step 4: Use the identity for \( x^4 + y^4 \)
We can use the identity:
\[ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 \]
Substituting \( x^2y^2 = 144 \):
\[ x^4 + y^4 = (x^2 + y^2)^2 - 2(144) \]
\[ x^4 + y^4 = (x^2 + y^2)^2 - 288 \]
Setting this equal to 337:
\[ (x^2 + y^2)^2 - 288 = 337 \]
Adding 288 to both sides:
\[ (x^2 + y^2)^2 = 337 + 288 \]
\[ (x^2 + y^2)^2 = 625 \]

### Step 5: Solve for \( x^2 + y^2 \)
Taking the square root of both sides:
\[ x^2 + y^2 = 25 \]

### Step 6: Use the identity to find \( x^2 - xy + y^2 \)
We know from algebra that:
\[ x^4 + y^4 + x^2y^2 = (x^2 + y^2)(x^2 - xy + y^2) \]
Substituting the known values:
\[ 481 = (25)(x^2 - xy + y^2) \]
Now, we need to isolate \( x^2 - xy + y^2 \):
\[ x^2 - xy + y^2 = \frac{481}{25} \]
Calculating the division:
\[ x^2 - xy + y^2 = 19.24 \]

### Step 7: Calculate \( x^2 - xy + y^2 \)
However, since we are looking for integer values, we need to use the identity correctly:
We can rearrange the equation:
\[ x^2 - xy + y^2 = x^2 + y^2 - xy \]
Substituting the known values:
\[ x^2 - xy + y^2 = 25 - 12 = 13 \]

### Final Answer
Thus, the value of \( x^2 - xy + y^2 \) is \( \boxed{13} \).