If x-7 divides `x^3-3k^3x^2-13x-7` , then k=

Options

A. 1.19
B. 1.34
C. 1.72
D. 4.63

Explanation

To solve the problem, we need to determine the value of \( k \) such that \( x - 7 \) divides the polynomial \( x^3 - 3k^3x^2 - 13x - 7 \). According to the polynomial remainder theorem, if \( x - 7 \) divides the polynomial, then substituting \( x = 7 \) into the polynomial should yield zero.

### Step-by-Step Solution:

1. **Substituting \( x = 7 \)**:
We start by substituting \( x = 7 \) into the polynomial:
\[ P(7) = 7^3 - 3k^3 \cdot 7^2 - 13 \cdot 7 - 7 \]

2. **Calculating \( P(7) \)**:
Now we calculate each term:
\[ 7^3 = 343 \]
\[ 7^2 = 49 \quad \text{so} \quad 3k^3 \cdot 7^2 = 3k^3 \cdot 49 = 147k^3 \]
\[ 13 \cdot 7 = 91 \]
Therefore, substituting these values into \( P(7) \):
\[ P(7) = 343 - 147k^3 - 91 - 7 \]

3. **Simplifying \( P(7) \)**:
Combine the constant terms:
\[ P(7) = 343 - 91 - 7 - 147k^3 = 343 - 98 - 147k^3 \]
\[ P(7) = 245 - 147k^3 \]

4. **Setting \( P(7) = 0 \)**:
Since \( x - 7 \) divides the polynomial, we set \( P(7) = 0 \):
\[ 245 - 147k^3 = 0 \]

5. **Solving for \( k^3 \)**:
Rearranging gives:
\[ 147k^3 = 245 \]
Dividing both sides by 147:
\[ k^3 = \frac{245}{147} \]

6. **Simplifying the fraction**:
We can simplify \( \frac{245}{147} \):
\[ \frac{245}{147} = \frac{35}{21} = \frac{5}{3} \]

7. **Finding \( k \)**:
Now we take the cube root:
\[ k = \sqrt[3]{\frac{5}{3}} = \left(\frac{5}{3}\right)^{1/3} \]

### Final Answer:
Thus, the value of \( k \) is:
\[ k = \left(\frac{5}{3}\right)^{1/3} \]