If (4a - 3b) = 1, ab = $$\frac{1}{2}$$ where a > 0 and b > 0, what is the value of (64a3 + 27b3)?
A. 15
B. 25
C. 30
D. 35
Answer: Option D
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
This Question Belongs to Arithmetic Ability >> Algebra
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