If (4x+2y)3 + (4x-2y)3 = 16 (Ax3 + Bry2), then what is the value of $\frac{1}{2}(\sqrt{A^2 + B^2})$ ?
Answer & explanation
Correct answer: option 3
(4x + 2y)3 + (4x - 2y)3 = 16 (Ax3 + Bxy2) .....(a)
a3 + b3 = (a + b) (a2 + b2 - ab)
(a + b)2 + (a - b)2 = 2 (a2 + b2)
a2 - b2 = (a + b) (a - b)
= (4x + 2y)3 + (4x - 2y)3 = (4x + 2y + 4x - 2y) [(4x + 2y)2 + (4x - 2y)2 - (4x + 2y) (4x - 2y)]
(4x + 2y)3 + (4x - 2y)3 = 8x [2 (16x2 + 4y2)] - (16x2 - 4y2)
(4x + 2y)3 + (4x - 2y)3 = 8x (32x2 + 8y2 - 16x2 + 4y2)
(4x + 2y)3 + (4x - 2y)3 = 8x (16x2 + 12y2)
(4x + 2y)3 + (4x - 2y)3 = 16 (8x3 + 6xy2) .....(b)
Justifying equations (1) and (2),
A = 8 and B = 6
$\frac{1}{2}(\sqrt{A^2 + B^2})$ = $\frac{1}{2}(\sqrt{8^2 + 6^2})$ = $\frac{1}{2}(\sqrt{100})$ = 5