If $16 x^2+y^2=48$ and $x y=2, x, y > 0$, then the value of $\left(64 x^3+y^3\right)$ is:
Answer & explanation
Correct answer: option 1
(a + b)3 = a3 + b3 + 3(ab)(a + b)
= 16x2 + y2 = 48
⇒ (4x)2 + y2 = 48
Add 8xy on both sides,
= (4x)2 + y2 + 8xy = 48 + 8xy
= (4x+y)2 = 48 + 16
= (4x+y)2 = 64
= 4x + y = 8
(a + b)3 = a3 + b3 + 3(ab)(a + b)
= (4x+y)3 = 83
= 64x3 + y3 + 3(4xy)(4x+y) = 512
= 64x3 + y3 + (3 × 8 × 8) = 512
= 64x3 + y3 = 512 – 192 = 320