Question : If $x^2+4y^2=17$ and $xy = 2$, where $x > 0, y > 0$, then what is the value of $x^3+8y^3$?
Option 1: 95
Option 2: 85
Option 3: 65
Option 4: 76
Answer (1)
18th Jan, 2024
Correct Answer: 65
Solution :
$x^2 + 4y^2 = 17$ and $xy = 2$, where $ x > 0, y > 0$
$⇒ xy = 2$
$⇒4xy = 8$
Now, $(x + 2y)^2= x^2 + 4y^2 + 4xy = 17 + 8 = 25$
$⇒ (x + 2y) = 5$
Also, $x^3 + 8y^3 = (x + 2y) (x^2 + 4y^2 - 2xy)$
$⇒ x^3 + 8y^2 = (5) × (17 - 4) = 65$
Hence, the correct answer is 65.
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