Directions (42-46) : In each question two equations numbered (I) and (II) are given. You have to solve both the equations and mark appropriate answer.

x³=729

y²-15y+54=0

To determine the correct relationship between $x$ and $y$, let us solve both given equations step-by-step.

Step 1: Solve Equation (I) for $x$

The first equation is:

$x^3=729$

Taking the cube root on both sides:

$x=\sqrt[3]{729}$

$x=9$

Step 2: Solve Equation (II) for $y$

The second equation is a quadratic equation:

$y^2-15y+54=0$

We split the middle term by finding two numbers that multiply to $54$ and add up to $-15$. These numbers are $-9$ and $-6$:

$y^2-9y-6y+54=0$

$y(y-9)-6(y-9)=0$

$(y-9)(y-6)=0$

Setting each factor equal to zero gives:

$y=9$ or $y=6$

Step 3: Compare the values of $x$ and $y$

We have $x=9$ and the possible values of $y$ are $9$ and $6$.
- When $y=9$, $x=y$.
- When $y=6$, $x>y$.

Combining both cases, we get:

$x\ge y$ (i.e., $x>=y$)

Hence, the correct answer is If x>=y.