If `8(a+b)^(3) + (a-b)^(3) = (3a+b)(A a^(2) + Bab + Cb^(2))`, then what is the value of `(A+B-C)` ?

Options

A. 2
B. 4
C. 10
D. 11

Answer

To solve the equation \( 8(a+b)^{3} + (a-b)^{3} = (3a+b)(A a^{2} + Bab + Cb^{2}) \), we will follow these steps:

### Step 1: Expand the left-hand side
We will use the identity for the sum of cubes:
\[ x^3 + y^3 = (x+y)(x^2 - xy + y^2) \]
Let \( x = 8(a+b) \) and \( y = (a-b) \).

First, we expand \( (a+b)^3 \) and \( (a-b)^3 \):
\[ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
\[ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \]

Now, substituting back:
\[ 8(a+b)^3 = 8(a^3 + 3a^2b + 3ab^2 + b^3) \]
\[ = 8a^3 + 24a^2b + 24ab^2 + 8b^3 \]

And for \( (a-b)^3 \):
\[ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \]

Combining these:
\[ 8(a+b)^3 + (a-b)^3 = (8a^3 + 24a^2b + 24ab^2 + 8b^3) + (a^3 - 3a^2b + 3ab^2 - b^3) \]
\[ = (8a^3 + a^3) + (24a^2b - 3a^2b) + (24ab^2 + 3ab^2) + (8b^3 - b^3) \]
\[ = 9a^3 + 21a^2b + 27ab^2 + 7b^3 \]

### Step 2: Expand the right-hand side
Now we expand the right-hand side:
\[ (3a+b)(Aa^2 + Bab + Cb^2) \]
Distributing:
\[ = 3a(Aa^2 + Bab + Cb^2) + b(Aa^2 + Bab + Cb^2) \]
\[ = 3Aa^3 + 3Bab^2 + 3Cb^2a + Aa^2b + Bab^2 + Cb^3 \]
\[ = 3Aa^3 + (3B + A)ab^2 + (3C + B) b^3 \]

### Step 3: Compare coefficients
Now we equate coefficients from both sides:
1. Coefficient of \( a^3 \): \( 9 = 3A \) → \( A = 3 \)
2. Coefficient of \( a^2b \): \( 21 = 3B + A \) → \( 21 = 3B + 3 \) → \( 3B = 18 \) → \( B = 6 \)
3. Coefficient of \( ab^2 \): \( 27 = 3C + B \) → \( 27 = 3C + 6 \) → \( 3C = 21 \) → \( C = 7 \)

### Step 4: Calculate \( A + B - C \)
Now we substitute \( A \), \( B \), and \( C \) into \( A + B - C \):
\[ A + B - C = 3 + 6 - 7 = 2 \]

### Final Answer
Thus, the value of \( A + B - C \) is \( \boxed{2} \).