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 :
Answer & explanation
Correct answer: option 1
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 = \(\sqrt {19}\)