Algebra

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,

\(x^3+y^3=216+27=243\)

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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 $x^3+y^3=6^3+3^3=243$.

x + y   = 9     square both  sides

x^2  + y^2  + 2xy   = 81

x^2  + y^2  + 2 (18)    =  81

x^2  +  y^2   +  36   =  81

x^2  +  y^2   =   45

And

x^3  + y^3    = 

 ( x + y)  ( x^2  - xy  +  y^2)   =

(x + y)  ( x^2 + y^2   -  xy)     =

 (9)  ( 45 - 18)  = 

(9) (27)    =

243