The real numbers x, y satisfy `x^(3)-3x^(2)+5x-17=0` and `y^(3)-3y^(2)+5y+11=0`, then `x+y` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equations given for \( x \) and \( y \), we have:

1. \( x^3 - 3x^2 + 5x - 17 = 0 \) (Equation 1)
2. \( y^3 - 3y^2 + 5y + 11 = 0 \) (Equation 2)

We want to find the value of \( x + y \).

### Step 1: Add the two equations

Let's add Equation 1 and Equation 2:

\[ (x^3 - 3x^2 + 5x - 17) + (y^3 - 3y^2 + 5y + 11) = 0 \]

This simplifies to:

\[ x^3 + y^3 - 3(x^2 + y^2) + 5(x + y) - 6 = 0 \]

### Step 2: Use the identities for sums of cubes and squares

Recall the identities:

- \( x^3 + y^3 = (x+y)(x^2 - xy + y^2) \)
- \( x^2 + y^2 = (x+y)^2 - 2xy \)

Let \( t = x + y \) and \( p = xy \). Then we can express \( x^3 + y^3 \) and \( x^2 + y^2 \) in terms of \( t \) and \( p \):

\[ x^3 + y^3 = t(t^2 - 3p) \]
\[ x^2 + y^2 = t^2 - 2p \]

### Step 3: Substitute into the equation

Substituting these into the equation we derived:

\[ t(t^2 - 3p) - 3(t^2 - 2p) + 5t - 6 = 0 \]

Expanding this gives:

\[ t^3 - 3tp - 3t^2 + 6p + 5t - 6 = 0 \]

Rearranging terms results in:

\[ t^3 - 3t^2 + (6 - 3p)t - 6 = 0 \]

### Step 4: Finding a specific value for \( t \)

Now, we can try specific values for \( t \). Let's test \( t = 2 \):

\[ 2^3 - 3(2^2) + (6 - 3p)(2) - 6 = 0 \]

Calculating this gives:

\[ 8 - 12 + (12 - 6p) - 6 = 0 \]
\[ -10 + 12 - 6p = 0 \]
\[ 2 - 6p = 0 \implies p = \frac{1}{3} \]

### Step 5: Conclusion

Since we found that \( t = 2 \) satisfies the equation, we conclude that:

\[ x + y = 2 \]

### Final Answer

Thus, the value of \( x + y \) is \( \boxed{2} \).