Question
If (x + 6y) = 8, and xy = 2, where x > 0, what is the
value of (x3 + 216y³)?Solution
xy = 2 ⇒ y = 2/x .....(1) Also, (x + 6y) = 8 ⇒ x + (6 × 2/x) = 8 ⇒ x + 12/x = 8 .....(2) ⇒ (x + 12/x)3 = 83 [Taking cube of both sides] ⇒ x3 + (12/x)3 + 3 × x × (12/x) (x + 12/x) 512 ⇒ x3 + (12/x)3 + 3 × 12 × 8 = 512 [āµ We get from equation (2), x + 12/x = 8] ⇒ x3 + (12/x)3 + 288 = 512 ⇒ x3 + (12/x)3 = 512 - 288 ⇒ x3 + [(2 × 6)/x]3 = 224 ⇒ x3 + (6y)3 = 224 [āµ y = 2/x] ⇒ x3 + 216y3 = 224 ∴ The value of x3 + 216y3 is 224
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