If x- y = 7 and ` x^(3) - y^(3) = 133. ` find :
`x^(2) + y^(2) `

Answer

To solve the problem, we are given two equations:

1. \( x - y = 7 \)
2. \( x^3 - y^3 = 133 \)

We need to find \( x^2 + y^2 \).

### Step 1: Use the identity for the difference of cubes
We know that:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
Substituting the value of \( x - y \):
\[ 133 = 7(x^2 + xy + y^2) \]

### Step 2: Solve for \( x^2 + xy + y^2 \)
Now, we can isolate \( x^2 + xy + y^2 \):
\[ x^2 + xy + y^2 = \frac{133}{7} = 19 \]

### Step 3: Use the identity for the square of a difference
We also know that:
\[ x^2 + y^2 = (x - y)^2 + 2xy \]
Substituting \( x - y = 7 \):
\[ x^2 + y^2 = 7^2 + 2xy = 49 + 2xy \]

### Step 4: Find \( xy \)
From the previous step, we have:
\[ x^2 + xy + y^2 = 19 \]
We can express \( x^2 + y^2 \) in terms of \( xy \):
\[ x^2 + y^2 = 19 - xy \]

### Step 5: Set up the equation
Now we have two expressions for \( x^2 + y^2 \):
1. \( x^2 + y^2 = 49 + 2xy \)
2. \( x^2 + y^2 = 19 - xy \)

Setting these equal to each other:
\[ 49 + 2xy = 19 - xy \]

### Step 6: Solve for \( xy \)
Rearranging gives:
\[ 49 + 3xy = 19 \]
\[ 3xy = 19 - 49 \]
\[ 3xy = -30 \]
\[ xy = -10 \]

### Step 7: Substitute \( xy \) back to find \( x^2 + y^2 \)
Now substitute \( xy = -10 \) back into the equation for \( x^2 + y^2 \):
\[ x^2 + y^2 = 49 + 2(-10) \]
\[ x^2 + y^2 = 49 - 20 \]
\[ x^2 + y^2 = 29 \]

### Final Answer
Thus, the value of \( x^2 + y^2 \) is:
\[ \boxed{29} \]