If 16x² + y² = 65over36 and xy = 1over18 , x, y < 0, then…

If 16x² + y² = $65\over36$ and xy = $1\over18$ , x, y < 0, then what is the value of (64x³ + y³)?

Show answer & explanation

Correct answer: D

Explanation

The correct answer is option 4 i.e. $19\over8$ .

16x² + y² = (4x)² + y² = 65/36

4xy = 4(1/18) = 2/9

(4x + y)² = 16x² + 8xy + y² = (65/36) + 2(2/9) = 9/4

Since x, y < 0 then 4x + y < 0

Then, 4x + y = -3/2

64x³ + y³ = (4x + y)(16x² - 4xy + y²)

= (-3/2) × (65/36 - 2/9)

= (-3/2) × (57/36)

= -19/8

SSC Selection Post 2022 PYQ — Question 64