In one method, you can use x and y, and figure out based on the expansion of (x+y)^3
But you can also notice that an x and y that satisfy \(x+y=9\) and \(xy=18\) is (6, 3)
Plugging these values in,
Solving for $x$ in $x+y=9$ yields $x=9-y$. Then plugging in this value of $x$ in the second expression yields $(9-y)y=18$. Solving for $y$ yields $y=3$ or $y=6$. Then we have the solutions $x=6$ and $x=3$ respecitvely. Now both sums are the same, so we have $x^3+y^3=6^3+3^3=243$.