This question picking numbers does not work very well

Better to use a general approach

We do know that \(x=y^3\)

Substitute in the two quantities

we do have

QA \(y^{{y{^3}}\)

QB \(y^{y+y+y}\)

We do have the same base so is the same to write

\(y^3>3y\)

\(y^3-3y>0\)

\(y(y^2-3)>0\)

Now

\(y>0\) which is false because we do know that \(y>1
\)

So we have to consider only \(y^2>3\)

Take the square root we do have that

\(y>1.7\)

We do know that \(y>1\)

However, we do have also \(y>1.7\)

Therefore y can be 1.1 or 2. Below 1.7 or greater than 1.7

The answer is D