If $x+y=9$ and $xy=10$, what is the value of $x^3+y^3$?

x^3 + y^3 =(x + y)(x^2 + xy + y^2)

= 9*((x + y)^2 - xy)

= 9*(9^2 - 10)

= 639.

Thanks for trying but the answer is 459

x= 9-y

xy would then be     (9-y) y  = 10       -y^2 +9y - 10 = 0     

                                                            y^2-9y+10 = 0

                                                           y = 9/2 +- sqrt41/ 2    

(x^3 +y^3  =    (  (9/2 + sqrt(41)/2 )^3  + (9/2 - sqrt(41)/2)^3 )   = 459   (I used WolframAlpha to do the multiplication)   

...and using Guest's original answer:   (with a slight correction ...see that ' - ' sign in the following line?)

x^3 + y^3 =(x + y)(x^2  -  xy + y^2)

               = (x+y)((x+y)^2 -3xy)

                     9 (9^2 -(3)10)  = 459

You could also tackle it this way:

(x + y)³ = x³  + 3x²y + 3xy² + y³ =  x³ + y³ + 3xy(x + y)

So  9³ =   x³ + y³ +3*10*9

x³ + y³ = 729 - 270 = 459

x^3 + y^3  =   ( x + y )  ( x^2 - xy + y^2)

And ( x + y) ^2   = 9^2   ⇒   x^2 + 2xy + y^2   = 81

And  since  xy  =  10  ....we have that

x^2 + 2(10) + y^2  = 81

x^2 + 20 + y^2   =81

x^2 + y^2  =  61

So

x^3 + y^3   =  ( x +y)  ( x^2 + y^2  - xy)  =  (9) ( 61  - 10)  =  9 * 51   =  459