If (4a - 3b) = 1, ab = $$\frac{1}{2}$$ where a > 0 and b > 0, what is the value of (64a³ + 27b³)?

Solution (By Examveda Team)

(4a - 3b) = 1, ab = $$\frac{1}{2}$$
Squaring both side
(4a - 3b)² = (1)²
16a² + 9b² - 2(4a)(3b) = 1
16a² + 9b² - 24ab = 1
16a² + 9b² - 24$$\left( {\frac{1}{2}} \right)$$ = 1
16a² + 9b² - 12 = 1
16a² + 9b² = 13
Adding 12 both side
16a² + 9b² + 12 = 13 + 12
(4a)² +(3b)² + 2(4a)(3b) = 25
(4a + 3b)² = 25
4a + 3b = 5
Cubing both side
64a³ + 27b³ = 125 - 3(4a)(3b)(4a + 3b)
64a³ + 27b³ = 125 - 90
64a³ + 27b³ = 35