Let `p(x)=0` be a polynomial equation of the least possible degree, with rational coefficients having `7 3+49 3` as one of its roots. Then product of all the roots of `p(x)=0` is `56` b. `63` c. `7` d. 49

Options

A. 7
B. 49
C. 56
D. 63

Answer

To solve the problem, we need to find the product of all the roots of the polynomial \( p(x) = 0 \) given that one of its roots is \( \sqrt[3]{7} + \sqrt[3]{49} \).

### Step-by-Step Solution:

1. **Identify the Root**:
Let \( x = \sqrt[3]{7} + \sqrt[3]{49} \).

2. **Cube Both Sides**:
To eliminate the cube roots, we cube both sides:
\[ x^3 = \left(\sqrt[3]{7} + \sqrt[3]{49}\right)^3 \]

3. **Apply the Cube Formula**:
Using the formula \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \), where \( a = \sqrt[3]{7} \) and \( b = \sqrt[3]{49} \):
\[ x^3 = \left(\sqrt[3]{7}\right)^3 + \left(\sqrt[3]{49}\right)^3 + 3\left(\sqrt[3]{7}\right)\left(\sqrt[3]{49}\right)\left(\sqrt[3]{7} + \sqrt[3]{49}\right) \]

4. **Calculate \( a^3 \) and \( b^3 \)**:
\[ a^3 = 7, \quad b^3 = 49 \]
Thus,
\[ x^3 = 7 + 49 + 3 \cdot \sqrt[3]{7} \cdot \sqrt[3]{49} \cdot x \]
Simplifying gives:
\[ x^3 = 56 + 3 \cdot \sqrt[3]{7 \cdot 49} \cdot x \]

5. **Simplify \( \sqrt[3]{7 \cdot 49} \)**:
Since \( 49 = 7^2 \):
\[ \sqrt[3]{7 \cdot 49} = \sqrt[3]{7^3} = 7 \]
Therefore, we have:
\[ x^3 = 56 + 21x \]

6. **Rearrange the Equation**:
Rearranging gives us the polynomial:
\[ x^3 - 21x - 56 = 0 \]

7. **Identify the Coefficients**:
In the polynomial \( x^3 - 21x - 56 = 0 \), the coefficients are:
- \( a = 1 \) (coefficient of \( x^3 \))
- \( b = 0 \) (coefficient of \( x^2 \))
- \( c = -21 \) (coefficient of \( x \))
- \( d = -56 \) (constant term)

8. **Calculate the Product of the Roots**:
The product of the roots of a cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \) is given by:
\[ \text{Product of roots} = -\frac{d}{a} = -\frac{-56}{1} = 56 \]

### Final Answer:
The product of all the roots of the polynomial \( p(x) = 0 \) is \( 56 \).