Solving the Algebraic Expression and Finding A, B, C

The problem asks us to analyze the given algebraic expression and its factored form to determine the values of constants A, B, and C, and then calculate the value of A - B - C.

The given equation is:

$\qquad (x + y)^3 + 27(x - y)^3 = (Ax - 2y)(Bx^2 + Cxy + 13y^2)$

We need to factor the left-hand side of the equation. The left side resembles the sum of cubes formula, which is $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.

Let $a = (x+y)$ and $b = \sqrt[3]{27}(x-y) = 3(x-y)$.

So, the expression is $a^3 + b^3$. Applying the sum of cubes formula:

First, calculate $(a+b)$:

$\qquad a + b = (x + y) + 3(x - y)$

$\qquad a + b = x + y + 3x - 3y$

$\qquad a + b = (x + 3x) + (y - 3y)$

$\qquad a + b = 4x - 2y$

Next, calculate $a^2$, $b^2$, and $ab$:

$\qquad a^2 = (x + y)^2 = x^2 + 2xy + y^2$

$\qquad b^2 = (3(x - y))^2 = 9(x - y)^2 = 9(x^2 - 2xy + y^2) = 9x^2 - 18xy + 9y^2$

$\qquad ab = (x + y)(3(x - y)) = 3(x + y)(x - y) = 3(x^2 - y^2) = 3x^2 - 3y^2$

Now, calculate $a^2 - ab + b^2$:

$\qquad a^2 - ab + b^2 = (x^2 + 2xy + y^2) - (3x^2 - 3y^2) + (9x^2 - 18xy + 9y^2)$

$\qquad a^2 - ab + b^2 = x^2 + 2xy + y^2 - 3x^2 + 3y^2 + 9x^2 - 18xy + 9y^2$

Combine like terms:

$\qquad x^2$ terms: $x^2 - 3x^2 + 9x^2 = (1 - 3 + 9)x^2 = 7x^2$

$\qquad xy$ terms: $2xy - 18xy = (2 - 18)xy = -16xy$

$\qquad y^2$ terms: $y^2 + 3y^2 + 9y^2 = (1 + 3 + 9)y^2 = 13y^2$

So, $a^2 - ab + b^2 = 7x^2 - 16xy + 13y^2$.

Now, substitute $(a+b)$ and $(a^2 - ab + b^2)$ back into the sum of cubes formula:

$\qquad (x + y)^3 + 27(x - y)^3 = (a+b)(a^2 - ab + b^2) = (4x - 2y)(7x^2 - 16xy + 13y^2)$

We are given that this factored form is equal to $(Ax - 2y)(Bx^2 + Cxy + 13y^2)$.

Comparing the two factored forms:

$(4x - 2y)(7x^2 - 16xy + 13y^2) = (Ax - 2y)(Bx^2 + Cxy + 13y^2)$

By comparing the coefficients of the corresponding terms, we can find the values of A, B, and C.

Comparing the first factors, $(4x - 2y)$ and $(Ax - 2y)$, we see that:

$\qquad A = 4$

Comparing the second factors, $(7x^2 - 16xy + 13y^2)$ and $(Bx^2 + Cxy + 13y^2)$, we see that:

• Coefficient of $x^2$: $B = 7$
• Coefficient of $xy$: $C = -16$
• Coefficient of $y^2$: $13 = 13$ (This confirms our calculation is consistent)

So, we have found the values:

• $A = 4$
• $B = 7$
• $C = -16$

The problem asks for the value of $A - B - C$.

$\qquad A - B - C = 4 - 7 - (-16)$

$\qquad A - B - C = 4 - 7 + 16$

$\qquad A - B - C = -3 + 16$

$\qquad A - B - C = 13$

Thus, the value of $A - B - C$ is 13.

Revision Table: Key Steps

Additional Information: Sum and Difference of Cubes

Factoring algebraic expressions is a fundamental skill in algebra. The sum and difference of cubes formulas are particularly useful for cubic expressions.

• Sum of Cubes: The formula for the sum of two cubes is $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. This formula was used in the problem above.
• Difference of Cubes: The formula for the difference of two cubes is $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$. Notice the sign changes compared to the sum of cubes formula. The first factor is $(a-b)$, and the second factor has all positive terms ($a^2 + ab + b^2$).

These formulas are derived from polynomial long division or by expanding the right-hand side. They are important for simplifying expressions, solving polynomial equations, and working with rational expressions.

In this problem, recognizing $27(x-y)^3$ as $(3(x-y))^3$ was key to applying the sum of cubes formula correctly.