If $$3\sqrt3 x^3 - 2\sqrt2 y^3 = (\sqrt3 x - \sqrt2y) (Ax^2 - Bxy + Cy^2)$$, then the value of $$(A^2 - B^2 + C^2)$$ is:

$$3\sqrt3 x^3 - 2\sqrt2 y^3 = (\sqrt3 x - \sqrt2y) (Ax^2 - Bxy + Cy^2)$$    Eq.(i)

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

As we know the above given formula. By this we will expand the given equation to obtain the values of A, B and C.

$$\left(\sqrt{3}x\right)^3+\left(-\sqrt{2}y\right)^3=(\sqrt{3}x+\left(-\sqrt{2}y\right))\ \times\ (\left(\sqrt{3}x\right)^2-\left(\left(\sqrt{3}x\ \times\ \left(-\sqrt{2}y\right)\right)\right)+\left(-\sqrt{2}y\right)^2)$$

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

Now compare the Eq.(i) with the above equation to obtain the values of A, B and C.

A = 3, B = $$-\sqrt{6}$$, C = 2

value of $$(A^2 - B^2 + C^2)$$ is = $$(3^2-(-\sqrt{6})^2+2^2)$$

= (9-6+4)

= 3+4

= 7