Find the cube root of `38-17sqrt5`

Options

A. `2+sqrt5`
B. `2-sqrt5`
C. `3-sqrt5`
D. `4-sqrt5`

Answer

To find the cube root of \( 38 - 17\sqrt{5} \), we can express it in the form \( a - b\sqrt{c} \) and use the identity for the cube of a binomial. Here are the steps to solve the problem:

### Step 1: Assume the form of the cube root
Let \( x = \sqrt[3]{38 - 17\sqrt{5}} \). We can assume that \( x \) can be expressed as \( a - b\sqrt{c} \), where \( a \) and \( b \) are rational numbers and \( c \) is a positive integer.

### Step 2: Cube the assumed form
Cubing both sides, we have:
\[ x^3 = (a - b\sqrt{c})^3 \]
Using the binomial expansion, we can expand this:
\[ x^3 = a^3 - 3a^2b\sqrt{c} + 3ab^2c - b^3c\sqrt{c} \]
This can be rearranged as:
\[ x^3 = (a^3 + 3ab^2c) - (3a^2b + b^3c)\sqrt{c} \]

### Step 3: Compare rational and irrational parts
From the equation \( x^3 = 38 - 17\sqrt{5} \), we can identify the rational and irrational parts:
- Rational part: \( a^3 + 3ab^2c = 38 \)
- Irrational part: \( -(3a^2b + b^3c) = -17 \)

### Step 4: Set up equations
From the irrational part, we can write:
\[ 3a^2b + b^3c = 17 \]
Assuming \( c = 5 \) (since we have \( \sqrt{5} \)), we can substitute \( c \) into the equations:
1. \( a^3 + 15ab^2 = 38 \)
2. \( 3a^2b + 5b^3 = 17 \)

### Step 5: Solve the equations
From the first equation, we can express \( a^3 \):
\[ a^3 = 38 - 15ab^2 \]
Substituting this into the second equation:
\[ 3a^2b + 5b^3 = 17 \]

### Step 6: Guess and check values for \( a \) and \( b \)
Let's try \( a = 2 \) and \( b = 1 \):
1. Check \( a^3 + 15ab^2 = 38 \):
\[ 2^3 + 15 \cdot 2 \cdot 1^2 = 8 + 30 = 38 \quad \text{(True)} \]
2. Check \( 3a^2b + 5b^3 = 17 \):
\[ 3 \cdot 2^2 \cdot 1 + 5 \cdot 1^3 = 12 + 5 = 17 \quad \text{(True)} \]

### Step 7: Conclusion
Since both equations hold true, we have \( a = 2 \) and \( b = 1 \). Therefore, the cube root of \( 38 - 17\sqrt{5} \) is:
\[ \sqrt[3]{38 - 17\sqrt{5}} = 2 - \sqrt{5} \]

### Final Answer
The cube root of \( 38 - 17\sqrt{5} \) is \( 2 - \sqrt{5} \).