Princ wrote:

What is the value of \(x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y\), Given x,y are positive integer.

Statement 1: \(x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24\)

Statement 2: \(x=y\)

gmatbusters Nice Approach

This is another approach.

We have to find value of \(x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y\), Given x,y are positive integer.

So we have to get individual value of x and y to get the definite value of expression .

Statement 1 : \(x^3 + y^ 3 + \frac{512}{x^3.y^3}= 24\)

Taking Arithmatic mean and Geometric mean of \(x^3\), \(y^ 3\), \(\frac{512}{x^3.y^3}\), we get

A.M=\(\frac{x^3 + y^3 +\frac{512}{x^3.y^3}}{3}\) ; G.M = \(\sqrt[3]{x^3*y^ 3*\frac{512}{x^3.y^3}}\) further simplifying, we get G.M =8.

As \(A.M\geq{G.M}\)

\(\frac{x^3 + y^3 +\frac{512}{x^3.y^3}}{3}\)≥8

\(x^3 + y^3 +\frac{512}{x^3.y^3}\)≥24

So Statement 1 is implying , that A.M is equal to G.M.This occurs when all terms are equal i.e

\(x^3\)=\(y^ 3\)=\(\frac{512}{x^3.y^3}\)

From \(y^ 3\)=\(\frac{512}{x^3.y^3}\) and substituting \(x^3\) with \(y^ 3\), we get \(y^9-512=0\)

\(y^9-2^9\)= \((y^3)^3-(2^3)^3\)\(=0\)

\((y^3)^3-(2^3)^3\)=\((y^3-2^3)(y^6+y^3.2^3+2^6)\)\(=0\),further expanding we get

\((y-2)(y^2+2.y+2^2)(y^6+y^3.2^3+2^6)=0\)

So \(y =2\), as \((y^2+2.y+2^2)(y^6+y^3.2^3+2^6)\) is +ve qty, cannot be equal to 0.

Similarly or using \(x^3=y^3\), we can get \(x=2\)

As we know both \(x\) and \(y\) , we can calculate expression \(x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y\)

Statement 1 is alone sufficient

Statement 2: \(x=y\)

As \(x\) and \(y\) are positive integer, They can take infinite different values even if \(x=y\).

Expression \(x^3.y^2 +\frac{2098}{x.y^2}+\frac{1}{x}-x^y\) will give infinite different value.

Statement 2 is not sufficient.

So OA should be A.

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