If `(x+y)^(3)+8(x-y)^(3)=(3x+Ay)(3x^(2)+Bxy+Cy^(2))`, then the value of `A+B+C` is:

Options

A. 3
B. 2
C. 4
D. 0

Answer

To solve the equation \( (x+y)^{3} + 8(x-y)^{3} = (3x + Ay)(3x^{2} + Bxy + Cy^{2}) \), we will follow these steps:

### Step 1: Expand the Left-Hand Side
We start by expanding the left-hand side of the equation:

\[ (x+y)^{3} + 8(x-y)^{3} \]

Using the binomial expansion, we have:

\[ (x+y)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3} \]

And for \( (x-y)^{3} \):

\[ (x-y)^{3} = x^{3} - 3x^{2}y + 3xy^{2} - y^{3} \]

Now, substituting this into our expression:

\[ 8(x-y)^{3} = 8(x^{3} - 3x^{2}y + 3xy^{2} - y^{3}) = 8x^{3} - 24x^{2}y + 24xy^{2} - 8y^{3} \]

Adding these two expansions together:

\[ (x+y)^{3} + 8(x-y)^{3} = (x^{3} + 3x^{2}y + 3xy^{2} + y^{3}) + (8x^{3} - 24x^{2}y + 24xy^{2} - 8y^{3}) \]

Combining like terms:

\[ = (1 + 8)x^{3} + (3 - 24)x^{2}y + (3 + 24)xy^{2} + (1 - 8)y^{3} \]

\[ = 9x^{3} - 21x^{2}y + 27xy^{2} - 7y^{3} \]

### Step 2: Expand the Right-Hand Side
Now we expand the right-hand side:

\[ (3x + Ay)(3x^{2} + Bxy + Cy^{2}) \]

Using the distributive property:

\[ = 3x(3x^{2}) + 3x(Bxy) + 3x(Cy^{2}) + Ay(3x^{2}) + Ay(Bxy) + Ay(Cy^{2}) \]

This simplifies to:

\[ = 9x^{3} + 3Bx^{2}y + 3Cx^{2}y + ABxy^{2} + ACy^{3} \]

Combining like terms gives us:

\[ = 9x^{3} + (3B + 3C)x^{2}y + ABxy^{2} + ACy^{3} \]

### Step 3: Equate Coefficients
Now we equate the coefficients from both sides:

From \( 9x^{3} \):
- Coefficient of \( x^{3} \): \( 9 = 9 \) (no new information)

From \( x^{2}y \):
- Coefficient of \( x^{2}y \): \( -21 = 3B + 3C \)

From \( xy^{2} \):
- Coefficient of \( xy^{2} \): \( 27 = AB \)

From \( y^{3} \):
- Coefficient of \( y^{3} \): \( -7 = AC \)

### Step 4: Solve the System of Equations
We have the following equations:

1. \( -21 = 3B + 3C \) (Divide by 3) → \( B + C = -7 \) (Equation 1)
2. \( 27 = AB \) (Equation 2)
3. \( -7 = AC \) (Equation 3)

From Equation 1, we can express \( C \) in terms of \( B \):

\[ C = -7 - B \]

Substituting \( C \) into Equation 3:

\[ -7 = A(-7 - B) \implies -7 = -7A - AB \]

Rearranging gives:

\[ 7A + AB = 7 \implies A(B + 7) = 7 \implies A = \frac{7}{B + 7} \]

Substituting \( A \) into Equation 2:

\[ 27 = \frac{7}{B + 7}B \implies 27(B + 7) = 7B \implies 27B + 189 = 7B \implies 20B = -189 \implies B = -\frac{189}{20} \]

Now substituting \( B \) back to find \( C \):

\[ C = -7 - \left(-\frac{189}{20}\right) = -\frac{140}{20} + \frac{189}{20} = \frac{49}{20} \]

Now substituting \( B \) into \( A(B + 7) = 7 \):

\[ A\left(-\frac{189}{20} + 7\right) = 7 \implies A\left(-\frac{189}{20} + \frac{140}{20}\right) = 7 \implies A\left(-\frac{49}{20}\right) = 7 \implies A = -\frac{140}{49} = -\frac{20}{7} \]

### Step 5: Calculate \( A + B + C \)
Now we can find \( A + B + C \):

\[ A + B + C = -\frac{20}{7} - \frac{189}{20} + \frac{49}{20} \]

Finding a common denominator (140):

\[ = -\frac{400}{140} - \frac{1323}{140} + \frac{343}{140} = -\frac{400 + 1323 - 343}{140} = -\frac{1380}{140} = -\frac{69}{7} \]

### Final Answer
Thus, the value of \( A + B + C \) is:

\[ \boxed{-\frac{69}{7}} \]