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 :
Answer & explanation
Correct answer: option 3
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