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

19

If $2\sqrt{2}x^3 - 3\sqrt{3}y^3 = (\sqrt{2}x- \sqrt{3}y) (Ax^2 - Bxy +Cy^2)$

 $(A^2 +B^2 +C^2)$

We know that =

a³ - b³ = ( a - b ) ( a² + b² + ab )

On comparing them we get = 

A = (\(\sqrt {2}\))² = 2

C = (\(\sqrt {3}\))² = 3

B = - (\(\sqrt {2}\)) × (\(\sqrt {3}\)) = -\(\sqrt {6}\)

$(A^2 +B^2 +C^2)$ = 2² +(-\(\sqrt {6}\))² + 3²

$(A^2 +B^2 +C^2)$  = 4 + 6 + + 9 = 19