The problem requires finding the sum of coefficients $A, B, C, D, E$ from the factorization of $8x^3 - 27y^3$.

Factorization Using Difference of Cubes

We use the algebraic identity for the difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

First, express $8x^3 - 27y^3$ in the form $a^3 - b^3$.

• $8x^3 = (2x)^3$, so $a = 2x$.
• $27y^3 = (3y)^3$, so $b = 3y$.

Applying the identity:

$8x^3 - 27y^3 = (2x)^3 - (3y)^3 = (2x - 3y)((2x)^2 + (2x)(3y) + (3y)^2)$

Simplify the second factor:

$8x^3 - 27y^3 = (2x - 3y)(4x^2 + 6xy + 9y^2)$

Identifying Coefficients

The problem states the factorization is $(Ax + By)(Cx^2 - Dy^2 + Exy)$.

Compare the terms from our derived factorization $(2x - 3y)(4x^2 + 6xy + 9y^2)$ with the given form:

• Comparing the first factor $(2x - 3y)$ with $(Ax + By)$:
  • $A = 2$
  • $B = -3$
• Comparing the second factor $(4x^2 + 6xy + 9y^2)$ with $(Cx^2 - Dy^2 + Exy)$:
  • For the $x^2$ term: $Cx^2 = 4x^2 \implies C = 4$.
  • For the $xy$ term: $Exy = 6xy \implies E = 6$.
  • For the $y^2$ term: $-Dy^2 = 9y^2 \implies -D = 9 \implies D = -9$.

Calculating the Coefficient Sum

Now, calculate the required sum $(A + B + C + D + E)$ using the identified coefficients:

$A + B + C + D + E = 2 + (-3) + 4 + (-9) + 6$

$A + B + C + D + E = 2 - 3 + 4 - 9 + 6$

$A + B + C + D + E = (2 + 4 + 6) + (-3 - 9)$

$A + B + C + D + E = 12 - 12$

$A + B + C + D + E = 0$