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28 Jan 2021, 01:15
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A
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Difficulty:
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36% (02:08) correct
64%
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wrong
based on 88
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What is the value of \(x^3y^2 +\frac{2098}{xy^2}+\frac{1}{x}-x^y\), given x, and y are positive integer?
(1) \(x^3 + y^ 3 + \frac{512}{x^3y^3}= 24\)
(2) \(x=y\)
Are You Up For the Challenge: 700 Level Questions
(1) \(x^3 + y^ 3 + \frac{512}{x^3y^3}= 24\)
(2) \(x=y\)
Are You Up For the Challenge: 700 Level Questions
29 Jan 2021, 12:42
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What is the value of x^3*y^2+2098/xy^2+1/x− x^y, given x, and y are positive integer?
Stat1: x^3+ y^3+ 512/(x^3*y^3)= 24
We know that, (a^+ b^3+ c^3) - 3abc = (a+b+c)*(a^2+b^2+c^2-ab-bc-ca)
where, a= x, b = y, c = 8/xy
So, given, (a^+ b^3+ c^3) = 3abc, which is possible, if (a+b+c)*(a^2+b^2+c^2-ab-bc-ca) = 0, or, (x+y+8/xy)* {x^2 + y^2+ 64/(x^2*y^2) - xy - 8/x - 8/y} =0,
as x and y are +ve integers, so, {x^2 +y^2+ 64/(x^2*y^2) -xy -8/x -8/y} =0 or,
x^2 + y^2+ 64/(x^2*y^2) = xy + 8/x + 8/y, which is only possible, x=y=2. Sufficient to get value of the req. expression.
Stat2: x=y; It is not sufficient.
So, I think A.
Stat1: x^3+ y^3+ 512/(x^3*y^3)= 24
We know that, (a^+ b^3+ c^3) - 3abc = (a+b+c)*(a^2+b^2+c^2-ab-bc-ca)
where, a= x, b = y, c = 8/xy
So, given, (a^+ b^3+ c^3) = 3abc, which is possible, if (a+b+c)*(a^2+b^2+c^2-ab-bc-ca) = 0, or, (x+y+8/xy)* {x^2 + y^2+ 64/(x^2*y^2) - xy - 8/x - 8/y} =0,
as x and y are +ve integers, so, {x^2 +y^2+ 64/(x^2*y^2) -xy -8/x -8/y} =0 or,
x^2 + y^2+ 64/(x^2*y^2) = xy + 8/x + 8/y, which is only possible, x=y=2. Sufficient to get value of the req. expression.
Stat2: x=y; It is not sufficient.
So, I think A.
12 Apr 2021, 23:18
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Statement 1 is enough.
The below screenshot shows how I approached the problem.
The below screenshot shows how I approached the problem.
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