To solve the problem, we need to find the equation whose roots are \( \alpha + \beta \), \( \beta + \gamma \), and \( \gamma + \alpha \) given that \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 - 7x + 6 = 0 \).
### Step 1: Identify the roots of the original polynomial
The given polynomial is:
\[
x^3 - 7x + 6 = 0
\]
We can factor this polynomial to find its roots.
### Step 2: Factor the polynomial
Using the Rational Root Theorem or trial and error, we can find that \( x = 1 \) is a root. We can then perform polynomial long division or synthetic division to factor the polynomial:
\[
x^3 - 7x + 6 = (x - 1)(x^2 + x - 6)
\]
Next, we can factor \( x^2 + x - 6 \):
\[
x^2 + x - 6 = (x - 2)(x + 3)
\]
Thus, the complete factorization is:
\[
x^3 - 7x + 6 = (x - 1)(x - 2)(x + 3)
\]
From this, we find the roots:
\[
\alpha = 1, \quad \beta = 2, \quad \gamma = -3
\]
### Step 3: Calculate the new roots
Now, we need to find the new roots:
1. \( \alpha + \beta = 1 + 2 = 3 \)
2. \( \beta + \gamma = 2 + (-3) = -1 \)
3. \( \gamma + \alpha = -3 + 1 = -2 \)
Thus, the new roots are \( 3, -1, -2 \).
### Step 4: Form the new polynomial
The polynomial whose roots are \( 3, -1, -2 \) can be formed using the fact that if \( r_1, r_2, r_3 \) are the roots, the polynomial can be expressed as:
\[
(x - r_1)(x - r_2)(x - r_3)
\]
Substituting the roots:
\[
(x - 3)(x + 1)(x + 2)
\]
### Step 5: Expand the polynomial
Now, we will expand this product:
1. First, multiply \( (x + 1)(x + 2) \):
\[
(x + 1)(x + 2) = x^2 + 3x + 2
\]
2. Now multiply this result by \( (x - 3) \):
\[
(x - 3)(x^2 + 3x + 2) = x^3 + 3x^2 + 2x - 3x^2 - 9x - 6
\]
Simplifying this gives:
\[
x^3 - 7x - 6
\]
### Final Result
Thus, the equation whose roots are \( \alpha + \beta, \beta + \gamma, \gamma + \alpha \) is:
\[
x^3 - 7x - 6 = 0
\]
### Conclusion
The correct answer is option number 4:
\[
x^3 - 7x - 6 = 0
\]
