If (4x + 2y)3 + (4x - 2y)3 = 16(Ax3 + Bxy2), then what is the value of $$\frac{1}{2}\left( {\sqrt {{A^2} + {B^2}} } \right)?$$
A. 8
B. 3
C. 5
D. 7
Answer: Option C
Solution (By Examveda Team)
(4x + 2y)3 + (4x - 2y)3 = 16(Ax3 + Bxy2)⇒ (4x)3 + (2y)3 + 3 × 4x × 2y(4x + 2y) + (4x)3 - (2y)3 - 3 × 4x × 2y(4x - 2y) = 16(Ax3 + Bxy2)
⇒ 2(4x)3 + 96x2y + 48xy2 - 96x2y + 48xy2 = 16(Ax3 + Bxy2)
⇒ 128x3 + 96xy2 = 16(Ax3 + Bxy2)
⇒ 16(8x3 + 6xy2) = 16(Ax3 + Bxy2)
On comparing on both side-
A = 8 & B = 6
Then,
$$\eqalign{ & \frac{1}{2}\left( {\sqrt {{A^2} + {B^2}} } \right) \cr & = \frac{1}{2}\left( {\sqrt {64 + 36} } \right) \cr & = \frac{1}{2} \times 10 \cr & = 5{\text{ Answer}} \cr} $$
This Question Belongs to Arithmetic Ability >> Algebra
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