If `40sqrt(5)x^(3) - 3sqrt(3)y^(3) = (2sqrt(5)x - sqrt(3)y) xx (Ax^(2) + Bxy + Cy^(2))`, then what is the value of `sqrt(B^(2) + C^(2) - A)` ?

Options

A. 11
B. 7
C. 8
D. 9

Answer

To solve the equation \( 40\sqrt{5}x^3 - 3\sqrt{3}y^3 = (2\sqrt{5}x - \sqrt{3}y)(Ax^2 + Bxy + Cy^2) \), we will follow these steps:

### Step 1: Recognize the structure of the equation
We can see that the left-hand side can be expressed in the form of a difference of cubes. The expression \( 40\sqrt{5}x^3 - 3\sqrt{3}y^3 \) can be rewritten using the formula for the difference of cubes:
\[ A^3 - B^3 = (A - B)(A^2 + AB + B^2) \]

### Step 2: Identify \( A \) and \( B \)
In our case:
- Let \( A = 2\sqrt{5}x \)
- Let \( B = \sqrt{3}y \)

### Step 3: Calculate \( A^3 \) and \( B^3 \)
Now we need to find \( A^3 \) and \( B^3 \):
- \( A^3 = (2\sqrt{5}x)^3 = 8 \cdot 5\sqrt{5}x^3 = 40\sqrt{5}x^3 \)
- \( B^3 = (\sqrt{3}y)^3 = 3\sqrt{3}y^3 \)

### Step 4: Use the difference of cubes formula
Now we can write:
\[ 40\sqrt{5}x^3 - 3\sqrt{3}y^3 = (2\sqrt{5}x - \sqrt{3}y)((2\sqrt{5}x)^2 + (2\sqrt{5}x)(\sqrt{3}y) + (\sqrt{3}y)^2) \]

### Step 5: Expand the second factor
Now we expand the second factor:
1. \( (2\sqrt{5}x)^2 = 4 \cdot 5 x^2 = 20x^2 \)
2. \( (2\sqrt{5}x)(\sqrt{3}y) = 2\sqrt{15}xy \)
3. \( (\sqrt{3}y)^2 = 3y^2 \)

Putting it all together, we have:
\[ 40\sqrt{5}x^3 - 3\sqrt{3}y^3 = (2\sqrt{5}x - \sqrt{3}y)(20x^2 + 2\sqrt{15}xy + 3y^2) \]

### Step 6: Identify coefficients \( A \), \( B \), and \( C \)
From the equation:
- \( A = 20 \)
- \( B = 2\sqrt{15} \)
- \( C = 3 \)

### Step 7: Calculate \( \sqrt{B^2 + C^2 - A} \)
Now we need to find \( \sqrt{B^2 + C^2 - A} \):
1. Calculate \( B^2 = (2\sqrt{15})^2 = 4 \cdot 15 = 60 \)
2. Calculate \( C^2 = 3^2 = 9 \)
3. Now substitute into the expression:
\[ B^2 + C^2 - A = 60 + 9 - 20 = 49 \]

### Step 8: Take the square root
Finally, we take the square root:
\[ \sqrt{49} = 7 \]

Thus, the value of \( \sqrt{B^2 + C^2 - A} \) is \( 7 \).