If (4x+2y)³ + (4x-2y)³ = 16 (Ax³ + Bry²), then what is the value of $\frac{1}{2}(\sqrt{A^2 + B^2})$ ?

5

(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