Question:
If 25x2 + y2 = 41 and xy = 4 where x, y > 0, then the value of (125x3 + y3) is :
If 25x2 + y2 = 41 and xy = 4 where x, y > 0, then the value of (125x3 + y3) is :
Updated On: Sep 10, 2024
- 189
- 1269
- 169
- 324
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The Correct Option is A
Solution and Explanation
Since, 25x2 + y2= 41
So, (5x)2 + y2= 41
Or, (5x + y)2 - 10xy = 41 (Since, (a + b)2 = a2 + b2 + 2ab)
Or, (5x + y)2 = 41 + 40 = 81
Or, 5x + y = 9
Cubing both sides, we get
125x3 + y3 + 15xy × (5x + y) = 729 {Since, (a + b)3 = a3 + b3 + 3ab(a + b)}
125x3 + y3 + 15 × 4 × 9 = 729
Or, 125x3 + y3= 729 - 540
Or, 125x3 + y3= 189
So, the correct option is (A) : 189.
So, (5x)2 + y2= 41
Or, (5x + y)2 - 10xy = 41 (Since, (a + b)2 = a2 + b2 + 2ab)
Or, (5x + y)2 = 41 + 40 = 81
Or, 5x + y = 9
Cubing both sides, we get
125x3 + y3 + 15xy × (5x + y) = 729 {Since, (a + b)3 = a3 + b3 + 3ab(a + b)}
125x3 + y3 + 15 × 4 × 9 = 729
Or, 125x3 + y3= 729 - 540
Or, 125x3 + y3= 189
So, the correct option is (A) : 189.
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