If `x+y=5, x^(3)+y^(3)=35`, then what is the positive difference between x and y?

Options

A. 0
B. 1
C. 5
D. 6

Answer

To find the positive difference between \( x \) and \( y \) given the equations \( x + y = 5 \) and \( x^3 + y^3 = 35 \), we can follow these steps:

### Step 1: Use the identity for the sum of cubes
We know that:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
We can rewrite \( x^2 - xy + y^2 \) in terms of \( x + y \) and \( xy \).

### Step 2: Substitute known values
From the problem, we have:
- \( x + y = 5 \)
- \( x^3 + y^3 = 35 \)

Substituting these values into the identity gives:
\[ 35 = 5(x^2 - xy + y^2) \]

### Step 3: Simplify the equation
Dividing both sides by 5:
\[ x^2 - xy + y^2 = 7 \]

### Step 4: Express \( x^2 + y^2 \) in terms of \( xy \)
We know that:
\[ x^2 + y^2 = (x + y)^2 - 2xy \]
Substituting \( x + y = 5 \):
\[ x^2 + y^2 = 5^2 - 2xy = 25 - 2xy \]

### Step 5: Substitute \( x^2 + y^2 \) into the equation
Now we can rewrite the equation:
\[ 25 - 2xy - xy = 7 \]
This simplifies to:
\[ 25 - 3xy = 7 \]

### Step 6: Solve for \( xy \)
Rearranging gives:
\[ 3xy = 25 - 7 \]
\[ 3xy = 18 \]
\[ xy = 6 \]

### Step 7: Use the values of \( x + y \) and \( xy \) to find \( x \) and \( y \)
We can now use the equations:
\[ t^2 - (x+y)t + xy = 0 \]
Substituting \( x + y = 5 \) and \( xy = 6 \):
\[ t^2 - 5t + 6 = 0 \]

### Step 8: Solve the quadratic equation
Factoring gives:
\[ (t - 2)(t - 3) = 0 \]
Thus, \( t = 2 \) or \( t = 3 \). Therefore, \( x \) and \( y \) are 2 and 3.

### Step 9: Find the positive difference
The positive difference between \( x \) and \( y \) is:
\[ |x - y| = |3 - 2| = 1 \]

### Final Answer
The positive difference between \( x \) and \( y \) is \( 1 \).

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