(1) \(\frac{(x^2y)^3}{(xy^3)^2} = 3\)

Simplifying,

\(\frac{(x^6*y^3)}{(x^2*y^6)} = 3\) => \(\frac{x^{(6-2)}}{y^{(6-3)}} = 3\) => \(\frac{(x^4)}{(y^3)} = 3\)

Because y can take any value, we cannot say anything clearly about the value of x(Insufficient)

(2) \(x^3y^{(–3)} = 1\)

This can also be written as \(\frac{(x^3)}{(y^3)} = 1\)

Because y can take any value, we cannot say anything clearly about the value of x(Insufficient)

On combining the information from both the statements,

If we divide equation in statement 2 from statement 1

\(\frac{{(x^4)}}{{(y^3)}}*\frac{{(y^3)}}{{(x^3)}} = \frac{3}{1}\)

This simplifies to give x = 3.

Therefore, the equation x>3 will always be false(Sufficient)

(Option C)
_________________

You've got what it takes, but it will take everything you've got