If $$(8x^3 - 27y^3) \div (2x — 3y) = (Ax^2 + Bxy + Cy^2)$$, then the value of (2A + B - C) is:

Given that,

$$(8x^3 - 27y^3) \div (2x — 3y) = (Ax^2 + Bxy + Cy^2)$$

But we know that,

$$a^3-b^3=(a-b)(a^2+b^2+ab)$$

So,  $$(8x^3 - 27y^3) \div (2x — 3y)=\dfrac{(2x)^3-(3y)^3}{2x-3y}$$

$$\dfrac{(2x)^3-(3y)^3}{2x-3y}=\dfrac{(2x-3y)((2x)^2+(2y)^2+2x\times 3y}{2x-3y}=4x^2+9y^2+6xy$$

Now, $$4x^2+9y^2+6xy=Ax^2+Cy^2+Bxy$$

Now, comparing both side,

$$A=4, B=6$$ and $$C=9$$

Now, substituting the values in the equation,

$$(2A + B - C)=(2\times 4+6-9)=5$$