SOLUTION: If x+y =3, then find x³+y³+10xy

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Question 1126980: If x+y =3, then find x³+y³+10xy
Found 2 solutions by rothauserc, ikleyn:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
Note that (a^3+b^3) = (a+b)(a^2−ab+b^2)
:
(x^3+y^3)+10xy = (x+y)(x^2-xy+y^2) +10xy =
:
3(x^2-xy+y^2) +10xy =
:
3x^2 -3xy +3y^2 +10xy =
:
3x^2 +7xy +3y^2
:

Answer by ikleyn(52852) About Me  (Show Source):
You can put this solution on YOUR website!
.
x%5E3+%2B+y%5E3+%2B+10xy = x%5E3+%2B+y%5E3+%2B+3%2Axy%2A3+%2B+xy = 


    in 3*xy*3 replace the second factor 3 by (x+y), based on what is given. Then you can continue this chain of equalities as 


= x%5E3+%2B+y%5E3+%2B+3xy%28x%2By%29 + xy = x%5E3+%2B+3x%5E2y+%2B+3xy%5E2+%2B+y%5E3 + xy = %28x%2By%29%5E3 + xy = 


    Replace (x+y) by 3 again, based on what is given. Then you can continue


= 3%5E3 + xy = 27 + xy.


That is all that can be done.

Completed and solved.