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 $\sqrt{A^2 +B^2+C^2}$ is :

$\sqrt{19}$

2√2x³ – 3√3y³ = (√2x – √3y) (Ax² + By² + Cxy)

= (√2x)³ – (√3y)³ = (√2x – √3y) (Ax² + By² + Cxy)

= (√2x – √3y) (2x² + 3y² + √6xy) = (√2x – √3y) (Ax² + By² + Cxy)

= (2x² + 3y² + √6 xy) = (Ax² + By² + Cxy)

On comparing

A = 2, B = 3 and C = √6

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

= 4 + 9 + 6 = 19 = \(\sqrt {19}\)