If (8x³ - 27y³) ÷ (2x - 3y) = (Ax² + Bxy + Cy²), then the value of (2A + B - C) is:

$$\eqalign{ & \left( {8{x^3} - 27{y^3}} \right) \div \left( {2x - 3y} \right) = A{x^2} + Bxy + C{y^2} \cr & \frac{{\left( {2x - 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right)}}{{\left( {2x - 3y} \right)}} = A{x^2} + Bxy + C{y^2} \cr & 4{x^2} + 6xy + 9{y^2} = A{x^2} + Bxy + C{y^2} \cr & {\text{Comparison both side}} \cr & A = 4,\,B = 6,\,C = 9 \cr & \left( {2A + B - C} \right) \cr & = \left( {2 \times 4 + 6 - 9} \right) \cr & = 5 \cr} $$