If $$(8x^3+27y^3)\div(2x+3y)=(Ax^2+Bxy+Cy^2)$$, then the value of $$(5A + 4B + 3C)$$ is:

Given,

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

$$\Rightarrow$$  $$\frac{\left(\left(2x\right)^3+\left(3y\right)^3\right)}{(2x+3y)}=(Ax^2+Bxy+Cy^2)$$

$$\Rightarrow$$  $$\frac{\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)}{(2x+3y)}=(Ax^2+Bxy+Cy^2)$$

$$\Rightarrow$$  $$4x^2-6xy+9y^2=\left(Ax^2+Bxy+Cy^2\right)$$

Comparing both sides we get,

A = 4, B = -6, C = 9

$$\therefore\ $$ $$5A+4B+3C=5\left(4\right)+4\left(-6\right)+3\left(9\right)$$

$$=20-24+27$$

$$=23$$

Hence, the correct answer is Option D